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Home » QnA » Line with slope greater than 1 passes A(4, 3) and intersects x–y=2 at B

Line with slope greater than 1 passes A(4, 3) and intersects x–y=2 at B

A line with a slope greater than 1 passes through point A(4, 3) and intersects the line x – y – 2 = 0 at point B. If the length of the line segment AB is \frac{{\sqrt {29} }}{3}, then B also lies on the line

  1. 2x + y = 9
  2. 3x – 2y = 7
  3. x + 2y = 6
  4. 2x – 3y = 3

Solution:

Firstly make the graph according to the question.

Use the formula to find the length of the line segment.

Then identify which line satisfy point B.

Step 1 of 3:

According to the question, we get

Step 2 of 3:

We have A(4, 3)

Let B(a, b) = B(a, a-2)

Also, given that length of line segment AB = \frac{{\sqrt {29} }}{3}

But we know that,

If A({x_1},{y_1})\& B({x_2},{y_2}) then length AB = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}}

\therefore AB = \sqrt {{{\left( {a - 4} \right)}^2} + {{\left( {a - 2 - 3} \right)}^2}}

\Rightarrow \sqrt {{{\left( {a - 4} \right)}^2} + {{\left( {a - 2 - 3} \right)}^2}} = \frac{{\sqrt {29} }}{3}

On squaring both sides

\begin{gathered} \Rightarrow {\left( {a - 4} \right)^2} + {\left( {a - 5} \right)^2} = \frac{{29}}{9} \hfill \\ \Rightarrow \left( {{a^2} - 8a + 16} \right) + \left( {{a^2} - 10a + 25} \right) = \frac{{29}}{9} \hfill \\ \Rightarrow 2{a^2} - 18a + 41 = \frac{{29}}{9} \hfill \\ \Rightarrow 9\left( {2{a^2} - 18a + 41} \right) = 29 \hfill \\ \Rightarrow 18{a^2} - 162a + 369 = 29 \hfill \\ \Rightarrow 18{a^2} - 162a + 369 - 29 = 0 \hfill \\ \Rightarrow 18{a^2} - 162a + 340 = 0 \hfill \\ \Rightarrow 2(9{a^2} - 81a + 170) = 0 \hfill \\ \Rightarrow 9{a^2} - 81a + 170 = 0 \hfill \\ \Rightarrow {a^2} - \frac{{81}}{9}a + \frac{{170}}{9} = 0 \hfill \\ \end{gathered}

Now, On the factorisation of the quadratic equation, we get

\Rightarrow \left( {a - \frac{{51}}{9}} \right)\left( {a - \frac{{10}}{3}} \right) = 0

\Rightarrow a = \frac{{51}}{9}or a = \frac{{10}}{3}

When a = \frac{{51}}{9} \Rightarrow b = a - 2

\Rightarrow b = \frac{{51}}{9} - 2

\Rightarrow b = \frac{{33}}{9}

Whena = \frac{{10}}{3} \Rightarrow b = a - 2

\begin{gathered} \Rightarrow b = \frac{{10}}{3} - 2 \hfill \\ \Rightarrow b = \frac{4}{3} \hfill \\ \end{gathered}

\therefore b = \left( {a,b} \right) \Rightarrow b = \left( {\frac{{51}}{9},\frac{{33}}{9}} \right)or\left( {\frac{{10}}{3},\frac{4}{3}} \right)

Step 3 of 3:

Line x + 2y = 6 is satisfy by \left( {\frac{{10}}{3},\frac{4}{3}} \right)

Final Answer:

Hence, Option (C) is correct

« General Solution of differential equation (x – y^2)dx +y(5x + y^2)dy
Mean and Variance of a binomial distribution are 24 and 128 »


Filed Under: QnA Tagged With: JEE Main 2022, July 25th Shift 1

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