Question 67

The graph of x + 2y = 3 and 3x - 2y = 1 meet the Y-axis at two points having distance

Solution

on Y axis, x=0

put x = 0 in x+2y = 3

2y = 3

$$ y = \frac{3}{2} $$

putting x=0 in 3x-2y = 1

-2y = 1

$$ \frac{-1}{2} $$

therefore points on Y-axis are

$$ (0,\frac{3}{2}) and (0,\frac{-1}{2}) $$

required distance = $$ \sqrt ((0-0)^2 + \sqrt (\frac{3}{2} + \frac{1}{2})^2 $$

$$ = \sqrt (0+4) $$ = 2 units

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