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The coordinates of the third vertex of an equilateral triangle whose two vertices are at (3, 4), (–2, 3) are ________.
Since, ΔABC is equilateral
∴ AB = AC ⇒ AB2 = AC2
⇒ (x – 3)2 + (y – 4)2
= (x + 2)2 + (y – 3)2
⇒ 5x + y – 6 = 0 ... (i)
Now, area of equilateral triangle =
∴ Area of ΔABC =
⇒ x – 5y = 13√3 − 17 ...(ii)
or x – 5y = −(13√3 + 17) ...(iii)
Solving (i) and (ii), we get,
Also, on solving (i) and (iii), we get
The vertices of a ΔABC are A(2,1), B(6, –2), C(8, 9). If AD is angle bisector, where D meets on BC, then coordinates of D are _______.
AD is the angle bisector of ∠BAC.
So, by the angle bisector theorem in ΔABC, we have
...(i)
Now, AB = 5 and AC = 10
∴ 1/2 = BD/DC [using (i)]
Thus, D divides BC in the ratio 1 : 2
∴
Find the coordinates of the point on X-axis which are equidistant from the points (–3, 4) and (2, 5).
Let P(x, 0) be the point on x-axis which is equidistant from points A(–3, 4) and B(2, 5)
Distance AP = Distance BP
⇒
⇒ (x + 3)2 + 16 = (x – 2)2 + 25
⇒ x2 + 9 + 6x + 16 = x2 + 4 – 4x + 25
⇒ 10x = 4 ⇒ x = 2/5
Co-ordinate of P (x, 0) is (2/5, 0).
In what ratio is the line segment joining the points (–3, 2) and (6, 1) is divided by Y-axis?
We know, co-ordinates on Y-axis is (0, y). Now, let P(0, y) divides the line segment joining the points A(–3, 2) and B(6, 1) in λ : 1.
Then,
∴ Required ratio = 1 : 2.
Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines)
According to question,
Distance covered from House to Bank
Distance covered from bank to school
Distance covered from school to office
∴ Total distance covered by Ayush = 5 + 10 + 12 = 27 km
Also, Distance covered directly from House to office
∴ Required distance = 27 – 11√5 = 2.40 km
If the centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at the origin, then a3 + b3 + c3 =
Since, centroid of triangle formed by (a, b), (b, c) and (c, a) is at the origin,
⇒ a + b + c = 0 ...(i)
Also, a3 + b3 + c3 – 3abc = 1/2 [a + b + c] × [(a – b)2 + (b – c)2 + (c – a)2]
⇒ a3 + b3 + c3 – 3abc = 0 (from (i))
⇒ a3 + b3 + c3 = 3abc
DIRECTION: Students of a school are standing in rows and columns in their playground for drill practice. A, B, C and D are the position of the four students as shown in the figure.
Q. If Mohit wants to stand in such a way that he is equidistance from each of the four students A, B, C and D then what are the coordinate of his position?
Mohit’s position will be (7, 5).
DIRECTION: Students of a school are standing in rows and columns in their playground for drill practice. A, B, C and D are the position of the four students as shown in the figure.
Q. What is the difference of distance between AC and AD?
From the given figure, Coordinates of A, B, C and D are respectively (3, 5), (7, 1), (11, 5) and (7, 9)
Distance from A to C
Distance from A to D
∴ Required difference = 8 – 4√2 = 2.34 units
To raise social awareness about t he hazards of smoking, a school decided to start “No smoking campaign”. A student is asked to prepare a campaign banner in the shape of a triangle shown in the figure. If cost of 1 cm2 of banner is ₹2, find the cost of the banner.
Area of banner
1/2= | [7(1− 1) + 1(1− 3) + 6(3 − 1)] = 5 cm2
Cost of 1 cm2 of banner = ₹2
∴ Cost of 5 cm2 of banner = 2 × 5 = ₹10
A well planned locality, has two straight roads perpendicular to each other. There are 5 lanes parallel to Road - I. Each lane has 8 houses as seen in figure. Chaitanya lives in the 6th house of the 5th lane and Hamida lives in the 2nd house of the 2nd lane. What will be the shortest distance between their houses?
According to question,
Coordinates of Chaitanya’s house is (6, 5)
Coordinates of Hamida’s house is (2, 2)
∴ Shortest distance between their houses
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