Question

the vertices of ΔABC are A(1,8),B(-2,4),C(8,-5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN and parallel to BC.

Answer 1

Given :

$A(1,8),B(-2,4),C(8,-5)\: \text{Are the vertices of }\Delta ABC$

$\text{M is the mid-point of AB}$

$\text{N is the mid-point of AC}$

To Find :

$\text{ Find the slope of MN}$

$\text{Verify that MN is Parallel to BC}$

Solution :

The given vertices of the triangle ABC are A(1,8),B(-2,4) and C(10,-5) M is the midpoint point of AB. N is midpoint of AC.

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A(1,8) }\put(0.5,-0.3){ \bf B(-2,4) }\put(5.2,-0.3){ \bf C(8,-5) }\qbezier(2,1.5)(2,1.5)(4,1.5)\put(1.6,1.5){\bf M}\put(4,1.5){\bf N}\put(1.9,1.42){ \bullet }\put(3.9,1.42){ \bullet }\end{picture}$

Slope of the line joining the points A(1,8) and B(-2,4) is

$M = (\frac{1-2}{2}, \frac{8+4}{2} )$ $(\because Mid-point\:formula = (\frac{x_2-x_1}{2},\frac{y_2-y_1}{2})$

$= (\frac{-1}{2},\frac{12}{2})\implies (\frac{-1}{2},6)$

Slope of the line joining the points A(1,8) and C(8,-5) is

$\therefore N = (\frac{1+8}{2},\frac{8-5}{2})$

$N = (\frac{9}{2},\frac{3}{2})$

Slope of the line joining the points $P(x_1,y_1)$ and $Q(x_2,x_1)$ is

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

Slope of the line joining the points

$M(\frac{-1}{2},6)$ and $N(\frac{9}{2},\frac{3}{2})$ is

$\text{Slope of MN}=\frac{\frac{3}{2}-6}{\frac{9}{2}-(-\frac{1}{2})}$

$\frac{\frac{3-12}{2} }{\frac{9}{2}+ \frac{1}{2} } = \frac{\frac{-9}{2} }{5} = \frac{-9}{10}$-------- (1)

Slope of the line joining the points B(-2,4) and (8,-2) is

Slope of BC = $\frac{-5-4}{8-(-2)} = \frac{-9}{8+2}=\frac{9}{10}$---------(2)

From Equation (1) and (2)

Slope of MN = Slope of BC

MN is parallel to BC

Final Answers :

Slope of MN is $\frac{-9}{10}$

From Equation (1) and (2) its verified that

Slope of MN = Slope of BC

If the slopes are equal so The lines will be parallel each other. Hence that

MN is parallel to BC