Test Index

KEAM 2013 Math Paper

Question : 54 of 120

Marks: +1, -0

(30, 40).

Solution:

Equation of straight line with slope 3 and passing through the point

(30,40)

(y-40)=3(x-30)

\Rightarrow\;\; y-40=3x-90

\Rightarrow\;\; 3x-y=50

\frac{x}{\frac{50}{3}} + \frac{y}{-50} = 1

.....(i)

and equation of straight line with slope 6 and passing through the point

(30,40)

(y-40)=6(x-30)

\Rightarrow\;\; y-40=6x-180

\Rightarrow

6x-y=140

\Rightarrow\; \frac{x}{\frac{70}{3}} + \frac{y}{-140} = 1

......(ii)

\therefore

The difference between the

y

-intercepts of the straight lines

=-50-(-140)=140-50=90

Go to Question:

Prev Question

Next Question