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Write down the equation of the line whose gradient is `-2/5 ` and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3:1. 

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To find the equation of the line with a gradient of \(-\frac{2}{5}\) that passes through point \(P\), which divides the line segment joining points \(A(4, -8)\) and \(B(12, 0)\) in the ratio \(3:1\), we will follow these steps: ### Step 1: Find the coordinates of point \(P\) using the section formula. The section formula states that if a point \(P(x_1, y_1)\) divides the line segment joining points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the ratio \(m:n\), then the coordinates of \(P\) can be calculated as follows: \[ x_1 = \frac{mx_2 + nx_1}{m+n} \] \[ y_1 = \frac{my_2 + ny_1}{m+n} \] Here, \(A(4, -8)\), \(B(12, 0)\), \(m = 3\), and \(n = 1\). Calculating the x-coordinate of \(P\): \[ x_1 = \frac{3 \cdot 12 + 1 \cdot 4}{3 + 1} = \frac{36 + 4}{4} = \frac{40}{4} = 10 \] Calculating the y-coordinate of \(P\): \[ y_1 = \frac{3 \cdot 0 + 1 \cdot (-8)}{3 + 1} = \frac{0 - 8}{4} = \frac{-8}{4} = -2 \] So, the coordinates of point \(P\) are \(P(10, -2)\). ### Step 2: Use the point-slope form to write the equation of the line. The point-slope form of the equation of a line is given by: \[ y - y_1 = m(x - x_1) \] Where \(m\) is the slope and \((x_1, y_1)\) are the coordinates of the point through which the line passes. Substituting \(m = -\frac{2}{5}\), \(x_1 = 10\), and \(y_1 = -2\): \[ y - (-2) = -\frac{2}{5}(x - 10) \] This simplifies to: \[ y + 2 = -\frac{2}{5}(x - 10) \] ### Step 3: Rearranging the equation to standard form. Now, we will simplify and rearrange the equation: \[ y + 2 = -\frac{2}{5}x + 4 \] Subtracting 2 from both sides: \[ y = -\frac{2}{5}x + 4 - 2 \] \[ y = -\frac{2}{5}x + 2 \] To convert this into standard form \(Ax + By + C = 0\), we can rearrange: \[ \frac{2}{5}x + y - 2 = 0 \] Multiplying through by 5 to eliminate the fraction: \[ 2x + 5y - 10 = 0 \] ### Final Equation of the Line Thus, the required equation of the line is: \[ 2x + 5y - 10 = 0 \] ---
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ICSE-EQUATION OF A LINE-EXERCISE 14(C)
  1. In DeltaABC, A = (3,5), B = (7, 8) and C = (1, -10). Find the equatio...

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  2. The following figure shows a parallelogram ABCD whose side AB is paral...

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  3. Find the equation of the straight line passing through origin and the ...

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  4. In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (...

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  5. A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) of triangle AB...

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  6. Find the equation of the perpendicular dropped from the point (-1, 2) ...

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  7. Find the equation of the line, whose : x-intercept = 5 and y-interce...

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  8. Find the equation of the line, whose : x-intercept = -4 and y-interc...

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  9. Find the equation of the line, whose : x-intercept = -8 and y-interc...

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  10. Find the equation of the line whose slope is -5/6 and x-intercept is...

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  11. Find the equation of the line with x-intercept 5 and a point on it (-3...

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  12. Find the equation of the line through (1, 3) and making an intercept o...

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  13. Find the equations of the lines passing through point (-2, 0) and equa...

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  14. The line through P(5, 3) intersects y-axis at Q. Write the slope ...

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  15. The line through P(5, 3) intersects y-axis at Q. Write the equati...

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  16. The line through P(5, 3) intersects y-axis at Q. Find the co-ordi...

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  17. Write down the equation of the line whose gradient is -2/5 and which ...

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  18. A (1, 4), B (3, 2) and C (7,5)  are vertices of a triangle ABC. Find :...

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  19. A (1, 4), B (3, 2) and C (7,5)  are vertices of a triangle ABC. Find :...

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  20. A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC....

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