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The straight lines x + y = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is :
- a)Isosceles
- b)Equilateral
- c)Right angled
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The straight lines x + y = 0, 3x + y – 4 = 0, x + 3y – 4 =...
X+y=0 .. (i)
3x+y−4=0 .. (ii)
x+3y−4=0 .... (iii)
Solving lines (i) and (ii), we get
−x=−3x+4⟹x=2⟹y=−2
∴(i) and (ii) intersect at A=(2,−2)
Solving lines (ii) and (iii), we get
−3x+4=34−x⟹−9x+12=4−x⟹x=1⟹y=1
∴(ii) and (iii) intersect at B=(1,1)
Solving lines (i) and (iii), we get
−x=34−x⟹−3x=4−x⟹x=−2⟹y=2
∴(i) and (iii) intersect at C=(−2,2)
So, AB,BC,AC form a triangle ABC
Now, AB=√(1−2)2+(1+2)2=10
BC=√(−2−1)2+(2−1)2=10
AC=√(−2−2)2+(2+2)2=32
∴AB=BC
Since, two sides of a triangle are equal then the triangle formed by A,B,C is isosceles triangle.
Hence, option A is correct.
3x+y−4=0 .. (ii)
x+3y−4=0 .... (iii)
Solving lines (i) and (ii), we get
−x=−3x+4⟹x=2⟹y=−2
∴(i) and (ii) intersect at A=(2,−2)
Solving lines (ii) and (iii), we get
−3x+4=34−x⟹−9x+12=4−x⟹x=1⟹y=1
∴(ii) and (iii) intersect at B=(1,1)
Solving lines (i) and (iii), we get
−x=34−x⟹−3x=4−x⟹x=−2⟹y=2
∴(i) and (iii) intersect at C=(−2,2)
So, AB,BC,AC form a triangle ABC
Now, AB=√(1−2)2+(1+2)2=10
BC=√(−2−1)2+(2−1)2=10
AC=√(−2−2)2+(2+2)2=32
∴AB=BC
Since, two sides of a triangle are equal then the triangle formed by A,B,C is isosceles triangle.
Hence, option A is correct.
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The straight lines x + y = 0, 3x + y – 4 = 0, x + 3y – 4 =...
The equation "x + y = 0" represents a straight line passing through the origin (0,0) with a slope of -1.
The equation "3x + y = 0" represents a straight line passing through the origin (0,0) with a slope of -3.
The equation "3x + y = 0" represents a straight line passing through the origin (0,0) with a slope of -3.
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The straight lines x + y = 0, 3x + y – 4 = 0, x + 3y – 4 = 0 form a triangle which is :a)Isoscelesb)Equilateralc)Right angledd)None of theseCorrect answer is option 'A'. Can you explain this answer?
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