Question

Find the slope of the line passing through the following points:
[i] (-3,2) and (1,4)
[ii] $\left( at_{1}^{2},2a{{t}_{1}} \right)$ and $\left( at_{2}^{2},2a{{t}_{2}} \right)$
[iii] (3,-5) and (1,2)

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
$m=\dfrac{4-2}{1-\left( -3 \right)}=\dfrac{2}{4}=\dfrac{1}{2}$
Hence the slope of the line is $\dfrac{1}{2}$
[ii] We have $A\equiv \left( at_{1}^{2},2a{{t}_{1}} \right)$ and $B\equiv \left( at_{2}^{2},2a{{t}_{2}} \right)$
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 ${{x}_{1}}=at_{1}^{2},{{x}_{2}}=at_{2}^{2},{{y}_{1}}=2a{{t}_{1}}$ and ${{y}_{2}}=2a{{t}_{2}}$
Hence, we have
$m=\dfrac{2a{{t}_{2}}-2a{{t}_{1}}}{at_{2}^{2}-at_{1}^{2}}=\dfrac{2a}{a}\left( \dfrac{{{t}_{2}}-{{t}_{1}}}{t_{2}^{2}-t_{1}^{2}} \right)$
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Hence, we have
$m=2\left( \dfrac{{{t}_{2}}-{{t}_{1}}}{\left( {{t}_{2}}-{{t}_{1}} \right)\left( {{t}_{2}}+{{t}_{1}} \right)} \right)=\dfrac{2}{{{t}_{1}}+{{t}_{2}}}$
Hence the slope of the line is $\dfrac{2}{{{t}_{1}}+{{t}_{2}}}$.
[iii] We have $A\equiv \left( 3,-5 \right)$ and $B\equiv \left( 1,2 \right)$
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 ${{x}_{1}}=3,{{x}_{2}}=1,{{y}_{1}}=-5$ and ${{y}_{2}}=2$
Hence, we have
$m=\dfrac{2-\left( -5 \right)}{1-3}=\dfrac{7}{-2}=-\dfrac{7}{2}$
Hence the slope of the line is $\dfrac{-7}{2}$

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 $-3m-m=2-4\Rightarrow m=\dfrac{-2}{-4}=\dfrac{1}{2}$
[ii] Let the equation of the line passing through the given points be y = mx+c
Since $\left( at_{1}^{2},2a{{t}_{1}} \right)$ lies on the line, we have
$\left( at_{1}^{2} \right)m+c=2a{{t}_{1}}$
Also since $\left( at_{2}^{2},2a{{t}_{2}} \right)$ lies on the line, we have
$\left( at_{2}^{2} \right)m+c=2a{{t}_{2}}$
Hence, we have $at_{1}^{2}m-at_{2}^{2}m=2a{{t}_{1}}-\left( 2a{{t}_{2}} \right)\Rightarrow m=\dfrac{2}{{{t}_{1}}+{{t}_{2}}}$.
[iii] Let the equation of the line passing through the given points be y = mx+c
Since (3-,5) lies on the line, we have
$3m+c=-5$
Also since (1,2) lies on the line, we have
$m+c=2$
Hence, we have $3m-m=-5-2\Rightarrow m=\dfrac{-7}{2}$