Question

23. If A(10,4), B(–4,9) and C(–2, –1) are the vertices of a triangle. Find the equation of

(i) AB (ii) the median through A

(iii) the altitude through B (iv) The perpendicular bisector of the side AB​

Answer 1

Answer:

This is your Correct Answer

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Answer 2

Step-by-step explanation:

Given:If A(10,4), B(–4,9) and C(–2, –1) are the vertices of the triangle.

To find: Find :-

(i) Equation of AB

(ii) the median through A

(iii) the altitude through B

(iv) The perpendicular bisector of the side AB

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) Equation of line AB.

Point A(10,4) and B(-4,9)

Equation of line AB:

$y-y_1=\frac{y_2 - y_1}{x_2 - x_1}(x-x_1)$

$y-4=\frac{9-4}{-4-10}(x-10)$

$y-4=\frac{-5}{14}(x-10)$

or

$14y - 56 = - 5x + 50$

or

$\bf \pink{5x - 14y = 106}$

(ii) Median through A:

Median passes through mid-point of opposite side BC

Step 1: find mid-point of BC

B(-4,9) and C(-2,-1)

Let the midpoint is E(x,y) and it is E(-3,4)

Step 2: Find equation of Median through A.

It passes through A(10,4) and E(-3,4)

Equation

$y -4 = \left(\frac{4-4}{10+3} \right)(x-10)$

$y - 4= 0$

$\bold{\green{y-4= 0}}$

(iii) Altitude Through B:

To find the equation of altitude through B(-5,9),find slope of altitude.

Step 1: Find slope of AC

A(10,4) and C(-2,-1)

$m = \frac{-1-4}{-2-10}$

$m = \frac{5}{12}$

Slope of altitude through B is -1/m

Slope of altitude through B is

$- \frac{12}{5}$

Step 2: Find equation of altitude through B.

it passes through B(-5,9), having slope -12/5

$y-9=\frac{-12}{5}(x+5)$

Simply

$5(y-9)=-12(x+5)$

$5y - 45 = - 12x - 60$

$\bf \red{12x + 5y + 15 = 0}$

(iv) The perpendicular bisector of the side AB.

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

Step 1: find mid-point of AB

A(10,4) and B(-4,9)

Let the midpoint is D(x,y)

$x = \frac{10 - 2}{2} \\ \\ x = 3$

$y = \frac{ 4 -9}{2} \\ \\ y = -2.5$

Step 2: Find equation of Perpendicular Bisector of AB.

It passes through C(-2,-1) and D(3,-2.5)

Equation

$y +1 = \left(\frac{-2.5 -1 }{3+2} \right)(x +2)$

$y +1 = \left(\frac{-3.5}{5} \right)(x +2)$

or

$y +1 = \left(\frac{-35}{50} \right)(x +2)$

$y +1 = \left(\frac{-7}{10} \right)(x +2)$

$10y + 10 = - 7x - 14$

or

$\bf \purple{7x + 10y + 24 = 0}$

Final answer:

(i) Equation of AB: $\bf \pink{5x - 14y = 106}$

(ii) The median through A: $\bold{\green{y-4= 0}}$

(iii) the altitude through B :$\bf \red{12x + 5y + 15 = 0}$

(iv) The perpendicular bisector of the side AB:$\bf \purple{7x + 10y + 24 = 0}$

Hope it helps you.

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