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Straight Lines and Pair of Straight Lines question

2019 · 9 Jan · Shift 2 · Q35
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Straight Lines and Pair of Straight Lines question

2019 · 9 Jan · Shift 2 · Q35

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1
Let the equations of two sides of a triangle be 3x −-− 2y + 6 = 0 and 4x + 5y −-− 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is :
  1. A
    122y −-− 26x −-− 1675 = 0
  2. B
    122y + 26x + 1675 = 0
  3. C
    26x + 61y + 1675 = 0
  4. D
    26x −-− 122y −-− 1675 = 0
View written solutionFree

Correct answer: D

  1. Given two sides of the triangle

Let L1:3x−2y+6=0L_1: 3x-2y+6=0L1​:3x−2y+6=0 L2:4x+5y−20=0L_2: 4x+5y-20=0L2​:4x+5y−20=0

The orthocentre is H(1,1).H(1,1).H(1,1).

We need the equation of the third side.

  1. Find the vertex formed by the given two sides

The intersection of L1L_1L1​ and L2L_2L2​ is one vertex, say AAA.

Solve 3x−2y+6=03x-2y+6=03x−2y+6=0 4x+5y−20=04x+5y-20=04x+5y−20=0

From the first equation, 3x−2y=−63x-2y=-63x−2y=−6

Multiply by 555: 15x−10y=−3015x-10y=-3015x−10y=−30

Multiply the second by 222: 8x+10y=408x+10y=408x+10y=40

Adding, 23x=10Rightarrowx=1023.23x=10 \\Rightarrow x=\frac{10}{23}.23x=10Rightarrowx=2310​.

Substitute into 4x+5y=204x+5y=204x+5y=20: 4(1023)+5y=204\left(\frac{10}{23}\right)+5y=204(2310​)+5y=20 4023+5y=20\frac{40}{23}+5y=202340​+5y=20 5y=420235y=\frac{420}{23}5y=23420​ y=8423.y=\frac{84}{23}.y=2384​.

So, A(1023,8423).A\left(\frac{10}{23},\frac{84}{23}\right).A(2310​,2384​).

  1. Altitude from vertex on side L1L_1L1​

If L1L_1L1​ is side ABABAB, then the opposite vertex is CCC, and the altitude from CCC passes through orthocentre HHH and is perpendicular to L1L_1L1​.

Equation of L1L_1L1​: 3x−2y+6=0⇒y=32x+33x-2y+6=0 \Rightarrow y=\frac{3}{2}x+33x−2y+6=0⇒y=23​x+3 So slope of L1L_1L1​ is m1=32.m_1=\frac{3}{2}.m1​=23​.

Hence slope of altitude from CCC is mCH=−23.m_{CH}=-\frac{2}{3}.mCH​=−32​.

Since it passes through (1,1)(1,1)(1,1), y−1=−23(x−1)y-1=-\frac{2}{3}(x-1)y−1=−32​(x−1) 3y−3=−2x+23y-3=-2x+23y−3=−2x+2 2x+3y−5=0.2x+3y-5=0.2x+3y−5=0.

So one altitude is h1:2x+3y−5=0.h_1: 2x+3y-5=0.h1​:2x+3y−5=0.

  1. Altitude from vertex on side L2L_2L2​

Equation of L2L_2L2​: 4x+5y−20=0⇒y=−45x+44x+5y-20=0 \Rightarrow y=-\frac{4}{5}x+44x+5y−20=0⇒y=−54​x+4 So slope of L2L_2L2​ is m2=−45.m_2=-\frac{4}{5}.m2​=−54​.

Hence slope of altitude from the opposite vertex is mBH=54.m_{BH}=\frac{5}{4}.mBH​=45​.

Passing through (1,1)(1,1)(1,1), y−1=54(x−1)y-1=\frac{5}{4}(x-1)y−1=45​(x−1) 4y−4=5x−54y-4=5x-54y−4=5x−5 5x−4y−1=0.5x-4y-1=0.5x−4y−1=0.

So the second altitude is h2:5x−4y−1=0.h_2: 5x-4y-1=0.h2​:5x−4y−1=0.

  1. Find the other two vertices using side-altitude intersections
  • Vertex BBB lies on side L1L_1L1​ and on altitude h2h_2h2​.
  • Vertex CCC lies on side L2L_2L2​ and on altitude h1h_1h1​.

Find BBB

Solve 3x−2y+6=03x-2y+6=03x−2y+6=0 5x−4y−1=05x-4y-1=05x−4y−1=0

Multiply the first by 222: 6x−4y+12=06x-4y+12=06x−4y+12=0 Subtract the second: (6x−4y+12)−(5x−4y−1)=0(6x-4y+12)-(5x-4y-1)=0(6x−4y+12)−(5x−4y−1)=0 x+13=0x+13=0x+13=0 x=−13.x=-13.x=−13.

Then 3(−13)−2y+6=03(-13)-2y+6=03(−13)−2y+6=0 −39−2y+6=0-39-2y+6=0−39−2y+6=0 −33−2y=0-33-2y=0−33−2y=0 y=−332.y=-\frac{33}{2}.y=−233​.

So, B(−13,−332).B\left(-13,-\frac{33}{2}\right).B(−13,−233​).

Find CCC

Solve 4x+5y−20=04x+5y-20=04x+5y−20=0 2x+3y−5=02x+3y-5=02x+3y−5=0

Multiply the second by 222: 4x+6y−10=04x+6y-10=04x+6y−10=0 Subtract the first: y+10=0y+10=0y+10=0 y=−10.y=-10.y=−10.

Then 2x+3(−10)−5=02x+3(-10)-5=02x+3(−10)−5=0 2x−35=02x-35=02x−35=0 x=352.x=\frac{35}{2}.x=235​.

So, C(352,−10).C\left(\frac{35}{2},-10\right).C(235​,−10).

  1. Equation of the third side BCBCBC

Now find the line through B(−13,−332),C(352,−10).B\left(-13,-\frac{33}{2}\right), \quad C\left(\frac{35}{2},-10\right).B(−13,−233​),C(235​,−10).

Slope of BCBCBC:

=\frac{\frac{13}{2}}{\frac{61}{2}}=\frac{13}{61}.$$ Using point-slope form through $B$: $$y+\frac{33}{2}=\frac{13}{61}(x+13).$$ Multiply by $122$: $$122y+2013=26x+338.$$ So, $$26x-122y-1675=0.$$ --- 7. **Compare with options** The equation obtained is $$26x-122y-1675=0,$$ which matches **Option D**. --- 8. **Comparison with stored correct answer** Stored correct answer: **D** Our derived answer: **D** So they agree.
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