Question

Find the slope of the lines:
[i] Passing through the points (3,-2) and (-1,4)
[ii] Passing through the points (3,-2) and (7,-2)

Answer 1

Hint: Use the fact that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ . Substitute the value of ${{x}_{1}},{{x}_{2}},{{y}_{1}},{{y}_{2}}$ in each case and hence find the slopes of the lines.
Alternatively, assume that the equation of the line is $y=mx+c$. Since the line passes through the points, the points satisfy the equation of the line. Hence form two linear equations in two variables m and c. Solve for m and c. The value of m gives the slope of the line.

Complete step-by-step answer:
[i] We have $A\equiv \left( 3,-2 \right)$ and $B\equiv \left( -1,4 \right)$
We know that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here ${{x}_{1}}=3,{{x}_{2}}=-1,{{y}_{1}}=-2$ and ${{y}_{2}}=4$
Hence, we have
$m=\dfrac{4+2}{-1-3}=\dfrac{6}{-4}=-\dfrac{3}{2}$
Hence the slope of the line is $-\dfrac{3}{2}$
[ii] We have $A\equiv \left( 3,-2 \right)$ and $B\equiv \left( 7,-2 \right)$
We know that the slope of the line joining the points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$.
Here ${{x}_{1}}=3,{{x}_{2}}=7,{{y}_{1}}=-2$ and ${{y}_{2}}=-2$
Hence, we have
$m=\dfrac{-2-\left( -2 \right)}{7-3}=\dfrac{0}{4}=0$
Hence the slope of the line is 0.
Note: Alternative solution:
[i] Let the equation of the line passing through the given points be y = mx+c
Since (3,-2) lies on the line, we have
$3m+c=-2$
Also since (-1,4) lies on the line, we have
$-m+c=4$
Hence, we have $3m+m=-2-4\Rightarrow m=\dfrac{-6}{4}=-\dfrac{3}{2}$
[ii] Let the equation of the line passing through the given points be y = mx+c
Since (3,-2) lies on the line, we have
$3m+c=-2$
Also since (7,-2) lies on the line, we have
$7m+c=-2$
Hence, we have $7m-3m=-2-\left( -2 \right)\Rightarrow m=0$.