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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएस.एस.एल.सी. (इंग्रजी माध्यम) इयत्ता १०

Find the equation of a straight line passing through the point P(−5, 2) and parallel to the line joining the points Q(3, −2) and R(−5, 4) - Mathematics

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प्रश्न

Find the equation of a straight line passing through the point P(−5, 2) and parallel to the line joining the points Q(3, −2) and R(−5, 4)

बेरीज

उत्तर

Slope of the line = `(y_2 - y_1)/(x_2 - x_1)`

Slope of the line QR = `(4 + 2)/(-5 - 3) = 6/(-8) = 3/(-4)`

⇒ `-3/4`

Slope of its parallel = `-3/4`

The given point is p(−5, 2)

Equation of the line is y – y1 = m(x – x1)

y – 2 = `-3/4(x + 5)`

4y – 8 = – 3x – 15

3x + 4y – 8 + 15 = 0

3x + 4y + 7 = 0

The equation of the line is 3x + 4y + 7 = 0

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General Form of a Straight Line
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Coordinate Geometry - Exercise 5.4 [पृष्ठ २३५]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 5 Coordinate Geometry
Exercise 5.4 | Q 5 | पृष्ठ २३५
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