हिंदी

Find the equation of a line passing through the point $$\mathrm{P}(–2, 1)$$ and parallel to the line joining the points $$\mathrm{A}(4, –3)$$ and $$\mathrm{B}(–1, 5)$$.

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प्रश्न

Find the equation of a line passing through the point $$\mathrm{P}(-2, 1)$$ and parallel to the line joining the points $$\mathrm{A}(4, -3)$$ and $$\mathrm{B}(-1, 5)$$.

योग
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उत्तर

Slope of line $$\mathrm{AB}$$ joining $$\mathrm{A}(4, -3)$$ and $$\mathrm{B}(-1, 5)$$: 

$$m = \frac{5 - (-3)}{-1 - 4}$$

$$= \frac{8}{-5}$$

$$= -\frac{8}{5}$$

Since the required line is parallel to $$\mathrm{AB}$$, its slope is also $$m = -\frac{8}{5}$$. 

The line passes through $$\mathrm{P}(-2, 1)$$: 

$$y - 1 = -\frac{8}{5}(x - (-2))$$ 

$$5(y - 1) = -8(x + 2)$$ 

$$5y - 5 = -8x - 16$$ 

$$8x + 5y + 11 = 0$$

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अध्याय 14: Equation of a Straight Line - EXERCISE 14B [पृष्ठ २३१]

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आर.एस. अग्रवाल Mathematics [English] Class 10 ICSE
अध्याय 14 Equation of a Straight Line
EXERCISE 14B | Q 14. | पृष्ठ २३१
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