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Find the equation of a line passing through the point $$(2, 3)$$ and intersecting the line $$2x – 3y = 6$$ on the $$y$$-axis. [Hint. 2x – 3у = 6 mеets y-axis at (0, –2). So, the required line passes

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Question

Find the equation of a line passing through the point $$(2, 3)$$ and intersecting the line $$2x - 3y = 6$$ on the $$y$$-axis.

[Hint. 2x – 3у = 6 mеets y-axis at (0, –2). So, the required line passes through (0, –2).]

Sum
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Solution

To find the point where $$2x - 3y = 6$$ meets the $$y$$-axis, put $$x = 0$$:

$$2(0) - 3y = 6$$

$$\implies -3y = 6$$

$$\implies y = -2$$

So, the line passes through $$(0, -2)$$.

The required line passes through $$(2, 3)$$ and $$(0, -2)$$.

Slope of the line: $$m = \frac{-2 - 3}{0 - 2}$$

$$= \frac{-5}{-2}$$

$$= \frac{5}{2}$$

Since the point $$(0, -2)$$ lies on the $$y$$-axis, the $$y$$-intercept is $$c = -2$$.

Using the slope-intercept form $$y = mx + c$$:

$$y = \frac{5}{2}x - 2$$

$$2y = 5x - 4$$

$$5x - 2y - 4 = 0$$

Thus, the equation of the line is $$5x - 2y - 4 = 0$$.

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Chapter 14: Equation of a Straight Line - EXERCISE 14A [Page 224]

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R.S. Aggarwal Mathematics [English] Class 10 ICSE
Chapter 14 Equation of a Straight Line
EXERCISE 14A | Q 19. | Page 224
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