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,
\[\begin{align}
  & \left( y-1 \right)=\left( \dfrac{1}{1+\dfrac{1}{2}} \right)\left( x+\dfrac{1}{2} \right) \\
 & \Rightarrow \left( y-1 \right)=\dfrac{2}{2+1}\left( x+\dfrac{1}{2} \right) \\
\end{align}\]
\[\left( y-1 \right)=\dfrac{2}{3}\left( x+\dfrac{1}{2} \right)\], Let us cross multiply
\[\begin{align}
  & 3\left( y-1 \right)=2\left( x+\dfrac{1}{2} \right) \\
 & \Rightarrow 3y-3=2x+1 \\
 & 2x-3y+4=0 \\
 & 2x-3y=-4 \\
\end{align}\]
Now let us divide throughout the equation by (-4).
\[\begin{align}
  & \dfrac{2x-3y}{-4}=\dfrac{-4}{-4} \\
 & \Rightarrow \dfrac{2x}{-4}-\dfrac{3y}{-4}=1 \\
 & \dfrac{-x}{2}+\dfrac{3y}{4}=1 \\
\end{align}\]
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.