Question:
A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2:1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be (α, β). Then, the value of \(7α + 3β\) is equal to ________ .
A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2:1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be (α, β). Then, the value of \(7α + 3β\) is equal to ________ .
Updated On: Feb 5, 2026
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Correct Answer: 31
Solution and Explanation
\(\frac {4}{5 - α} = \frac {3}{α - 2}\)
\(⇒ 4α - 8 = 15 - 3α\)
\(α = \frac {23}{7}\)
\(A = ( \frac {23}{7},0)\ Q = (5,4)\)
\(R = (\frac {10 + \frac {23}{7}}{3} , \frac 83 )\)
\(= ( \frac {31}{7} , \frac 83 )\)
Bisector of angle PAQ is \(X = \frac {23}{7}\).
\(⇒ M = ( \frac {23}{7} , \frac 83 )\)
So, \(7α + 3β = 31\)
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Concepts Used:
Tangents and Normals
- A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
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- Slope of tangent (at x=x0) m=dy/dx||x=x0
- A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got
m×n = -1
- The normal to a given curve y = f(x) at a point x = x0
- The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m
Diagram Explaining Tangents and Normal:
