Question

Consider the line $4x-3y-10=0$
(a) Prove that $(4,2)$ is a point on this line. Find another point on this line.
(b) Find the slope of this line.
(c) Write the equation of the line with same slope and passing through the point $(3,5)$.

Answer 1

Equation of line is $4x-3y-10=0$
(a) To check whether $(4,2)$ is a point on this line, we put $x=4$ and $y=2$ in the equation of line.
i.e. $4\left(4\right)-3\left(2\right)-10=16-6-10=0$.
Hence, $(4,2)$ is a point on this line
Let the x-coordinate of another point be $1$.
Putting $x=1$ in the equation of line, we get
$4\left(1\right)-3y-10=0$
i.e. $-3y-6=0$
i.e. $-3y=6$
i.e. $y=-2$
Thus, $(1,-2)$ is another point on this line.
(b) Let the co-ordinates of two points on this line be $(x_1,y_1)$ and $(x_2,y_2)$.
Then, we have
$4x_1-3y_1-10=0$ and $4x_2-3y_2-10=0$
Thus, we have
$(4x_2-3y_1-10)-(4x_2-3y_2-10)=0$
$\Rightarrow 4(x_1-x_2)-3(y_1-y_2)=0$
$\Rightarrow 4(x_1-x_2)=3(y_1-y_2)$
$\Rightarrow \frac{y_1-y_2}{x_1-x_2}=\frac{4}{3}$
Thus, the slope of the line is $\frac{4}{3}$.
(c) Slope $=\frac{4}{3}$ and point is $(3,5)$
For any point $(x,y)$ on this line, we have
$\frac{y-5}{x-3}=\frac{4}{3}$
$\Rightarrow 3y-15=4x-12$
$\Rightarrow 4x-3y+3=0$
Hence, equation of line is $4x-3y+3=0$