Question

If A(-5,-7), B(13,2) and C (-5,6) are the vertices of the triangle then Find :-

(i) Altitude Through A
(ii) Perpendicular Bisector through B
(iii) Median Through C

Chapter Name :- Pair Of Straight lines

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Answer 1

Answer:

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Answer 2

Given:If A(-5,-7), B(13,2) and C (-5,6) are the vertices of the triangle.

To find: Find :-

(i) Altitude Through A

(ii) Perpendicular Bisector through B

(iii) Median Through C

Solution:

Tip:

*Slope of two perpendicular lines follow the relationship $\bold{m_1m_2 = - 1}$

*Equation of line having slope m and passes through (x1,y1)

$\bold{y-y_1=m(x-x_1)}$

*Slope of line passing through two points (x1,y1) and (x2,y2)

$\bold{m = \frac{y_2 - y_1}{x_2 - x_1}}$

(i) Altitude Through A:

To find the equation of altitude through A(-5,-7),find slope of altitude.

Step 1: Find slope of BC

B(13,2) and C(-5,6)

$m = \frac{6-2}{-5-13}$

$m = \frac{4}{-18}$

$m = \frac{-2}{9}$

Slope of altitude through A is -1/m

Slope of altitude through A is 9/2

Step 2: Find equation of altitude through A.

it passes through A(-5,-7), having slope 9/2

$y+7=\frac{9}{2}(x+5)$

Simply

$2(y+7)=9(x+5)$

$2y+14=9x+45$

$\bold{\red{9x-2y+31=0}}$

(ii) Perpendicular Bisector through B:

Perpendicular Bisector passes through mid-point of opposite side AC.

Step 1: find mid-point of AC

A(-5,-7) and C(-5,6)

Let the midpoint is D(x,y)

$x = \frac{ - 5 - 5}{2} \\ \\ x = - 5$

$y = \frac{ - 7 + 6}{2} \\ \\ y = - 0.5$

Step 2: Find equation of Perpendicular Bisector through B.

It passes through B(13,2) and D(-5,-0.5)

Equation

$y + 0.5 = \left(\frac{2 + 0.5 }{13 + 5} \right)(x + 5)$

$y + 0.5 = \left(\frac{2.5 }{18} \right)(x + 5)$

$18y + 9 = 2.5x + 12.5 \\ \\ 2.5x - 18y + 3.5 = 0 \\ \\ or \\ \\ \bold{\orange{25x - 180y + 35 = 0}}$

(iii) Median through C:

Median passes through mid-point of opposite side AB

Step 1: find mid-point of AB

A(-5,-7) and C(13,2)

Let the midpoint is E(x,y) and it is E(4,-2.5)

Step 2: Find equation of Median through C.

It passes through C(-5,6) and D(4,-2.5)

Equation

$y + 2.5 = \left(\frac{6 + 2.5 }{-5-4} \right)(x -4)$

$y + 2.5 = \left(\frac{8.5 }{-9} \right)(x -4)$

$9y + 22.5 = -8.5x +34$

$\bold{\green{8.5x+9y =11.5}}$

Final answer:

(i) Altitude Through A: $9x-2y+31=0$

(ii) Perpendicular Bisector through B: $25x-180y+35=0$

(iii) Median Through C: $8.5x+9y =11.5$

Hope it helps you.

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