Question

Find the slope of the line which passes through the points
[i] (0,0) and (4,-2) [ii] (0,-3) and (2,1)

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.

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