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Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 PDF Download

Question 8: Find the area of the triangle PQR with Q(3, 2) and the mid-points of the sides through Q being (2, −1) and (1, 2).

Answer : Let P(x₁, y₁), Q(3, 2) and R(x₂, y₂) be the vertices of the ∆PQR.
Suppose S(2, −1) and T(1, 2) be the mid-points of sides QR and PQ, respectively.

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Using mid-point formula, we have

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

⇒x1=−1, y1=2

Coordinates of P = (−1, 2)

Also,

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Coordinates of R = (1, −4)

So, P(−1, 2), Q(3, 2) and R(1, −4) are the vertices of ∆PQR.

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

=12 square units

Hence, the area of the triangle is 12 square units.

Question 9: If P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.

Answer : It is given that P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2) are the vertices of a quadrilateral PQRS.

Area of the quadrilateral PQRS = Area of ∆PQR + Area of ∆PRS

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

=1/2|−5[−6−(−3)]+(−4)[−3−(−3)]+2[−3−(−6)]|

=1/2|15+0+6|

= 21/2 square units

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

= 35/2 square units

∴ Area of the quadrilateral PQRS = Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 square units

Hence, the area of the given quarilateral is 28 square units.

Question 10: If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.

Answer : It is given that A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD.

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10
Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

=1/2
|−3[−7−(−8)]+(−2)(−8−5)+1[5−(−7)]|

=1/2|−3+26+12|

=35/2 square units

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

=1/2|−3(−8−3)+1(3−5)+6[5−(−8)]|

=1/2|33−2+78|

= 109/2 square units

∴ Area of the quadrilateral ABCD = Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 $\frac{35}{2} + \frac{109}{2} = \frac{144}{2} = 72$ square units

Hence, the area of the given quadrilateral is 72 square units.

Question 11: For what value of a the point (a, 1), (1, −1), and (11, 4) are collinear?

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(a, 1), B(1, −1) and C(11, 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

Δ=1/2 |(x1y2+x2y3+x3y1)−(x2y1+x3y2+x1y3)|

0=1/2 |(a×−1+1×4+11×1)−(1×1+11×−1+a×4)|

0=1/2 | (−a+4+11)−(1−11+4a)|

0=1/2|(−a+15)−(−10+4a)|

0=1/2|−a+15+10−4a|

0=1/2|−5a+25|

0=−5a+25

5a=25

a=5

Hence the value of ‘a’ for which the given points are collinear is Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 12:Prove that the points (a, b), (a₁, b₁) and (a −a₁, b −b₁) are collinear if ab₁ = a₁b.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Δ=1/2|(x1y2+x2y3+x3y1)−(x2y1+x3y2+x1y3)|∆=12x1y2+x2y3+x3y1-x2y1+x3y2+x1y3

If three points are collinear the area encompassed by them is equal to 0.

The three given points are Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 , and. If they are collinear then the area enclosed by them should be 0.

Δ=1/2|(ab1+a1(b−b1)+(a−a1)b)−(a1b+(a−a1)b1+a(b−b1))|

0=1/2|(ab1+a1b−a1b1+ab−a1b)−(a1b+ab1−a1b1+ab−ab1)|

0=1/2|ab1+a1b−a1b1+ab−a1b−a1b−ab1+a1b1−ab+ab1|

0=ab1−a1b

ab1=a1b

Hence we have proved that for the given conditions to be satisfied we need to have Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 13: If the vertices of a triangle are (1, −3), (4, p) and (−9, 7) and its area is 15 sq. units, find the value(s) of p.

Answer : Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC.
Here, x₁ = 1, y₁ = −3; x₂ = 4, y₂ = p and x₃ = −9, y₃ = 7
ar(∆ABC) = 15 square units

$\Rightarrow \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| = 15 \Rightarrow \frac{1}{2} |1 (p - 7) + 4 [7 - (- 3)] + (- 9) (- 3 - p)| = 15 \Rightarrow \frac{1}{2} |p - 7 + 40 + 27 + 9 p| = 15 \Rightarrow |10 p + 60| = 30$

$\Rightarrow 10 p + 60 = 30$ or $10 p + 60 = - 30$
$\Rightarrow 10 p = - 30$ or $10 p = - 90$
$\Rightarrow p = - 3$ or $p = - 9$
Hence, the value of p is −3 or −9.

Question 14: If (x, y) be on the line joining the two points (1, −3) and (−4, 2) , prove that x + y + 2= 0.

Answer : Since the point (x, y) lie on the line joining the points (1, −3) and (−4, 2); the area of triangle formed by these points is 0.

That is,

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Thus, the result is proved.

Question 15: Find the value of k if points A(k, 3), B(6, −2) and C(−3, 4) are collinear.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Δ=1/2|(x1y2+x2y3+x3y1)−(x2y1+x3y2+x1y3)|∆=12x1y2+x2y3+x3y1-x2y1+x3y2+x1y3

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(k, 3), B(6, −2) and C(−3, 4). It is also said that they are collinear and hence the area enclosed by them should be 0.

Δ=1/2|(k(−2)+6×4+(−3)×3)−(6×3+(−3)(−2)+k×4)| 0=1/2|(−2k+24−9)−(18+6+4k)| 0=1/2|−2k+15−24−4k| 0=1/2|−6k−9| 0=−6k−9 k=−9/6 =−3/2

Hence the value of ‘k’ for which the given points are collinear is Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 16: Find the value of k, if the points A(7, −2), B (5, 1) and C (3, 2k) are collinear.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

$∆ = \frac{1}{2} |(x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{1}) - (x_{2} y_{1} + x_{3} y_{2} + x_{1} y_{3})|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(7, −2), B(5, 1) and C(3, 2k). It is also said that they are collinear and hence the area enclosed by them should be 0.

$∆ = \frac{1}{2} |(7 \times 1 + 5 \times 2 k + 3 \times - 2) - (5 \times - 2 + 3 \times 1 + 7 \times 2 k)| 0 = \frac{1}{2} |(7 + 10 k - 6) - (- 10 + 3 + 14 k)| 0 = \frac{1}{2} |- 4 k + 8| 0 = - 4 k + 8 k = 2$

Hence the value of ‘k’ for which the given points are collinear is Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 17: If the point P (m, 3) lies on the line segment joining the points $A (- \frac{2}{5}, 6)$ and B (2, 8), find the value of m.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

$∆ = \frac{1}{2} |(x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{1}) - (x_{2} y_{1} + x_{3} y_{2} + x_{1} y_{3})|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point P(m,3) lies on the line segment joining the points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 and B(2,8). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

0=2m+8 m=−4

Hence the value of ‘m’ for which the given condition is satisfied is Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 18: If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that x + y = a + b.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

$∆ = \frac{1}{2} |(x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{1}) - (x_{2} y_{1} + x_{3} y_{2} + x_{1} y_{3})|$

If three points are collinear the area encompassed by them is equal to 0.

It is said that the point R(x, y) lies on the line segment joining the points P(a, b) and Q(b, a). Hence we understand that these three points are collinear. So the area enclosed by them should be 0.

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

0=ay+xa+b2−xb−by−a2 a2−b2=ax+ay−bx−by (a+b)(a−b)=(a−b)(x+y) (a+b)=(x+y)

Hence under the given conditions we have proved that Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 19: Find the value of k, if the points A (8, 1) B(3, −4) and C(2, k) are collinear.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

$∆ = \frac{1}{2} |(x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{1}) - (x_{2} y_{1} + x_{3} y_{2} + x_{1} y_{3})|$

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(8,1), B(3,−4) and C(2,k). It is also said that they are collinear and hence the area enclosed by them should be 0.

Δ=1/2|(8×−4+3×k+2×1)−(3×1+2×−4+8×k)| 0=1/2|(−32+3k+2)−(3−8+8k)| 0=1/2|−25−5k|k=−5

Hence the value of ‘k’ for which the given points are collinear is Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 20:Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Δ=1/2|(x1y2+x2y3+x3y1)−(x2y1+x3y2+x1y3)|∆=12x1y2+x2y3+x3y1-x2y1+x3y2+x1y3

The three given points are A(a,2a), B(−2,6) and C(3,1). It is also said that the area enclosed by them is 10 square units. Substituting these values in the above mentioned formula we have,

$∆ = \frac{1}{2} |(a \times 6 + (- 2) \times 1 + 3 \times 2 a) - ((- 2) \times 2 a + 3 \times 6 + a \times 1)| 10 = \frac{1}{2} |(6 a - 2 + 6 a) - (- 4 a + 18 + a)| 10 = \frac{1}{2} |15 a - 20| 20 = |15 a - 20| 4 = |3 a - 4|$

We have Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 . Hence either

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Hence the values of ‘a’ which satisfies the given conditions are Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

Question 21: Prove that the points (a, 0), (0, b) and (1, 1) are collinear if $\frac{1}{a} + \frac{1}{b} = 1$ .

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

We know area of triangle formed by three points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 is given by

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(a,0), B(0,b) and C(1,1).

$A = \frac{1}{2} |a (b - 1) + 1 (0 - b)| = \frac{1}{2} |a b - a - b|$

It is given that Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

So we have,

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Using this in the previously arrived equation for area we have,

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Since the area enclosed by the three points is equal to 0, the three points need to be Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 .

**Question 22: The point A divides the join of P (−5, 1) and Q(3, 5) in the ratio k:1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C(7, −2) is equal to 2 units.**

Answer : GIVEN: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio k: 1

Coordinates of point B (1, 5) and C (7, −2)

TO FIND: The value of k

PROOF: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio k: 1

So the coordinates of A are Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

We know area of triangle formed by three points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 is given by $∆ = \frac{1}{2} |(x_{1} y_{2} + x_{2} y_{3} + x_{3} y_{1}) - (x_{2} y_{1} + x_{3} y_{2} + x_{1} y_{3})|$

Now Area of ΔABC= 2 sq units.

Taking three points A Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 , B (1, 5) and C (7, −2)

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Hence Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Question 23: The area of a triangle is 5. Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y = x + 3. Find the third vertex.

Answer : GIVEN: The area of triangle is 5.Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y= x+3

TO FIND: The third vertex.

PROOF: Let the third vertex be (x, y)

Now

Taking three points(x, y), (2, 1) and (3, −2)

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Also it is given the third vertex lies on y = x+3

Substituting the value in equation (1) and (2) we get

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Hence the coordinates of Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 and

Question 24: If , prove that the points (a, a²), (b, b²), (c, c²) can never be collinear.

Answer : GIVEN: If

TO PROVE: that the points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 can never be collinear.

PROOF:

We know three points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 are collinear when

Now taking three points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Also it is given that

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Hence area of triangle made by these points is never zero. Hence given points are never collinear.

Question 25: Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that find x.

Answer : GIVEN: four points A (6, 3), B (−3, 5) C (4, −2) and D(x, 3x) such that

TO FIND: the value of x

PROOF:

We know area of the triangles formed by three points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 is given by

Now

Area of triangle DBC taking D(x, 3x), B (−3, 5), C (4, −2)

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Area of triangle ABC taking, A (6, 3), B (−3, 5), C (4, −2)

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Also it is given that

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Substituting the values from (1) and (2) we get

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Question 26: If three points (x₁, y₁) (x₂, y₂), (x₃, y₃) lie on the same line, prove that

Answer : GIVEN: If three points lie on the same line

TO PROVE: Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

PROOF:

We know that three points Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10 are collinear if

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents, Videos & Tests for Class 10

Hence proved.

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Question 8: Find the area of the triangle PQR with Q(3, 2) and the mid-points of the sides through Q being (2, −1) and (1, 2).

Answer : Let P(x1, y1), Q(3, 2) and R(x2, y2) be the vertices of the ∆PQR.Suppose S(2, −1) and T(1, 2) be the mid-points of sides QR and PQ, respectively.

Question 9: If P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.

Answer : It is given that P(−5, −3), Q(−4, −6), R(2, −3) and S(1, 2) are the vertices of a quadrilateral PQRS.

Question 10: If A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.

Answer : It is given that A(−3, 5), B(−2, −7), C(1, −8) and D(6, 3) are the vertices of a quadrilateral ABCD.

Question 11: For what value of a the point (a, 1), (1, −1), and (11, 4) are collinear?

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 12:Prove that the points (a, b), (a1, b1) and (a −a1, b −b1) are collinear if ab1 = a1b.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 13: If the vertices of a triangle are (1, −3), (4, p) and (−9, 7) and its area is 15 sq. units, find the value(s) of p.

Answer : Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC.Here, x1 = 1, y1 = −3; x2 = 4, y2 = p and x3 = −9, y3 = 7ar(∆ABC) = 15 square units

Question 14: If (x, y) be on the line joining the two points (1, −3) and (−4, 2) , prove that x + y + 2= 0.

Answer : Since the point (x, y) lie on the line joining the points (1, −3) and (−4, 2); the area of triangle formed by these points is 0.

Question 15: Find the value of k if points A(k, 3), B(6, −2) and C(−3, 4) are collinear.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 16: Find the value of k, if the points A(7, −2), B (5, 1) and C (3, 2k) are collinear.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 17: If the point P (m, 3) lies on the line segment joining the points A(−2/5,6) A-25,6 and B (2, 8), find the value of m.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 18: If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that x + y = a + b.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 19: Find the value of k, if the points A (8, 1) B(3, −4) and C(2, k) are collinear.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 20:Find the value of a for which the area of the triangle formed by the points A(a, 2a), B(−2, 6) and C(3, 1) is 10 square units.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 21: Prove that the points (a, 0), (0, b) and (1, 1) are collinear if =11a+1b=1.

Answer : The formula for the area ‘A’ encompassed by three points, and is given by the formula,

Question 22: The point A divides the join of P (−5, 1) and Q(3, 5) in the ratio k:1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C(7, −2) is equal to 2 units.

Answer : GIVEN: point A divides the line segment joining P (−5, 1) and Q (3, −5) in the ratio k: 1

Question 23: The area of a triangle is 5. Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y = x + 3. Find the third vertex.

Answer : GIVEN: The area of triangle is 5.Two of its vertices are (2, 1) and (3, −2). The third vertex lies on y= x+3

Question 24: If , prove that the points (a, a2), (b, b2), (c, c2) can never be collinear.

Answer : GIVEN: If

Question 25: Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that find x.

Answer : GIVEN: four points A (6, 3), B (−3, 5) C (4, −2) and D(x, 3x) such that

Question 26: If three points (x1, y1) (x2, y2), (x3, y3) lie on the same line, prove that

Answer : GIVEN: If three points lie on the same line

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Answer : Let P(x₁, y₁), Q(3, 2) and R(x₂, y₂) be the vertices of the ∆PQR.
Suppose S(2, −1) and T(1, 2) be the mid-points of sides QR and PQ, respectively.

Question 12:Prove that the points (a, b), (a₁, b₁) and (a −a₁, b −b₁) are collinear if ab₁ = a₁b.

Answer : Let A(1, −3), B(4, p) and C(−9, 7) be the vertices of the ∆ABC.
Here, x₁ = 1, y₁ = −3; x₂ = 4, y₂ = p and x₃ = −9, y₃ = 7
ar(∆ABC) = 15 square units

Question 17: If the point P (m, 3) lies on the line segment joining the points $A (- \frac{2}{5}, 6)$ and B (2, 8), find the value of m.

Question 21: Prove that the points (a, 0), (0, b) and (1, 1) are collinear if $\frac{1}{a} + \frac{1}{b} = 1$ .

**Question 22: The point A divides the join of P (−5, 1) and Q(3, 5) in the ratio k:1. Find the two values of k for which the area of ΔABC where B is (1, 5) and C(7, −2) is equal to 2 units.**

Question 24: If , prove that the points (a, a²), (b, b²), (c, c²) can never be collinear.

Question 26: If three points (x₁, y₁) (x₂, y₂), (x₃, y₃) lie on the same line, prove that

Coordinate Geometry Exercise 14.1 (Part-13) | Extra Documents

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