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Question: Find the slope of the line which passes through the points [i] (2,5) and (-4,-4) [ii] (-2,3) and ...

Find the slope of the line which passes through the points
[i] (2,5) and (-4,-4) [ii] (-2,3) and (4,-6)

Explanation

Solution

Hint: Use the fact that the slope of the line joining the points A(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) is given by m=y2y1x2x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} . Substitute the value of x1,x2,y1,y2{{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+cy=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(2,5)A\equiv \left( 2,5 \right) and B(4,4)B\equiv \left( -4,-4 \right)
We know that the slope of the line joining the points A(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) is given by m=y2y1x2x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}.
Here x1=2,x2=4,y1=5{{x}_{1}}=2,{{x}_{2}}=-4,{{y}_{1}}=5 and y2=4{{y}_{2}}=-4
Hence, we have
m=4542=96=32m=\dfrac{-4-5}{-4-2}=\dfrac{-9}{-6}=\dfrac{3}{2}
Hence the slope of the line is 32\dfrac{3}{2}
[ii] We have A(2,3)A\equiv \left( -2,3 \right) and B(4,6)B\equiv \left( 4,-6 \right)
We know that the slope of the line joining the points A(x1,y1)A\left( {{x}_{1}},{{y}_{1}} \right) and B(x2,y2)B\left( {{x}_{2}},{{y}_{2}} \right) is given by m=y2y1x2x1m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}.
Here x1=2,x2=4,y1=3{{x}_{1}}=-2,{{x}_{2}}=4,{{y}_{1}}=3 and y2=6{{y}_{2}}=-6
Hence, we have
m=634(2)=96=32m=\dfrac{-6-3}{4-\left( -2 \right)}=\dfrac{-9}{6}=\dfrac{-3}{2}
Hence the slope of the line is 32\dfrac{-3}{2}.
Note: Alternative solution:
[i] Let the equation of the line passing through the given points be y = mx+c
Since (2,5) lies on the line, we have
2m+c=52m+c=5
Also since (-4,-4) lies on the line, we have
4m+c=4-4m+c=-4
Hence, we have 2m+4m=5+4m=96=322m+4m=5+4\Rightarrow m=\dfrac{9}{6}=\dfrac{3}{2}
[ii] Let the equation of the line passing through the given points be y = mx+c
Since (-2,3) lies on the line, we have
2m+c=3-2m+c=3
Also since (4,-6) lies on the line, we have
4m+c=64m+c=-6
Hence, we have 4m+2m=63m=324m+2m=-6-3\Rightarrow m=\dfrac{-3}{2}.