Question

If a straight line passes through the points $\left( \dfrac{-1}{2},1 \right)$ and (1, 2) then its x – intercept is:
(a) -2
(b) -1
(c) 1
(d) 0

Answer 1

Hint:Using a two point formula of line, substitute the point $\left( \dfrac{-1}{2},1 \right)$ and (1, 2) and find the equation. Compare the equation with the equation of line which cuts off intercepts at a and b i.e., $\dfrac{x}{a}+\dfrac{y}{b}=1$. Here a is the x – intercept.

Complete step-by-step answer:
It is said that a straight line passes through the points $\left( \dfrac{-1}{2},1 \right)$ and (1, 2). Now we know the two points form of line formula as,
$\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
Let us arrange the above equation as,
$y-{{y}_{1}}=\left( \dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\left( x-{{x}_{1}} \right)... (1)$
Now, put $\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{-1}{2},1 \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)=\left( 1,2 \right)$.
Let us substitute the values of $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in equation (1).
Hence (1) becomes,
$\left( y-1 \right)=\left[ \dfrac{2-1}{1-\left( -\dfrac{1}{2} \right)} \right]\left( x-\left( -\dfrac{1}{2} \right) \right)$
Let us simplify the above expression,

& \left( y-1 \right)=\left( \dfrac{1}{1+\dfrac{1}{2}} \right)\left( x+\dfrac{1}{2} \right) \\\ & \Rightarrow \left( y-1 \right)=\dfrac{2}{2+1}\left( x+\dfrac{1}{2} \right) \\\ \end{aligned}$$ $$\left( y-1 \right)=\dfrac{2}{3}\left( x+\dfrac{1}{2} \right)$$, Let us cross multiply $$\begin{aligned} & 3\left( y-1 \right)=2\left( x+\dfrac{1}{2} \right) \\\ & \Rightarrow 3y-3=2x+1 \\\ & 2x-3y+4=0 \\\ & 2x-3y=-4 \\\ \end{aligned}$$ Now let us divide throughout the equation by (-4). $$\begin{aligned} & \dfrac{2x-3y}{-4}=\dfrac{-4}{-4} \\\ & \Rightarrow \dfrac{2x}{-4}-\dfrac{3y}{-4}=1 \\\ & \dfrac{-x}{2}+\dfrac{3y}{4}=1 \\\ \end{aligned}$$ We can also write the above equation as, $$\dfrac{x}{-2}+\dfrac{y}{\left( \dfrac{4}{3} \right)}=1.......(2)$$ Now this expression is of the form, $$\dfrac{x}{a}+\dfrac{y}{b}=1$$ - (3) Hence here ‘a’ is the x – intercept from the above expression. Thus by comparing equation (2) and (3) we get, $$a=-2$$ Thus, a is the x – intercept here i.e. a = - 2. $$\therefore $$ We got the x – intercept as – 2. Hence, option (a) is the correct answer. Note: The equation $$\dfrac{x}{a}+\dfrac{y}{b}=1$$, is the equation of the line in the intercept form, which will satisfy the coordinates at any point in a line. If you need the y – intercept then take the value of b.