1

### JEE Main 2018 (Online) 15th April Evening Slot

The sides of a rhombus ABCD are parallel to the lines, x $-$ y + 2 = 0 and 7x $-$ y + 3 = 0. If the diagonals of the rhombus intersect P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the coordinate of A is :
A
${5 \over 2}$
B
${7 \over 4}$
C
2
D
${7 \over 2}$

## Explanation

Let the coordinate A be (0, c)

Equations of the given lines are

x $-$ y + 2 = 0 and 7x $-$ y + 3 = 0

We know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; y = x + 2 and y = 7x + 3

$\therefore\,\,\,$ equation of angle bisectors is given as :

${{x - y + 2} \over {\sqrt 2 }} = \pm {{7x - y + 3} \over {5\sqrt 2 }}$

5x $-$ 5y + 10 = $\pm$ (7x $-$ y + 3)

$\therefore\,\,\,$ Parallel equations of the diagonals are 2x + 4y $-$ 7 = 0

and 12x $-$ 6y + 13 = 0

$\therefore\,\,\,$ slopes of diagonals are ${{ - 1} \over 2}$ and 2.

Now, slope of the diagonal from A(0, c) and passing through P(1, 2) is (2 $-$ c)

$\therefore\,\,\,$ 2 $-$ c = 2 $\Rightarrow$ c = 0 (not possible)

$\therefore $$\,\,\, 2 - c = {{ - 1} \over 2} \Rightarrow c = {5 \over 2} \therefore\,\,\, Coordinate of A is {5 \over 2}. 2 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Evening Slot The foot of the perpendicular drawn from the origin, on the line, 3x + y = \lambda (\lambda \ne 0) is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is : A 1 : 3 B 3 : 1 C 1 : 9 D 9 : 1 ## Explanation Equation of the line, which is perpendicular to the line, 3x + y = \lambda (\lambda \ne 0) and passing through origin , is given by {{x - 0} \over 3} = {{y - 0} \over 1} = r For foot of perpendicular r = {{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)} \over {{3^2} + {1^2}}} = {\lambda \over {10}} So, foot of perpendicular P = \left( {{{3\lambda } \over {10}},{\lambda \over {10}}} \right) Given the line meets X-axis where y = 0, so 3x + 0 = \lambda \Rightarrow x = {\lambda \over 3} Hence, coordinates of A = \left( {{\lambda \over 3},0} \right) and meets Y-axis at B = (0, \lambda ) So, BP = \sqrt {{{\left( {{{3\lambda } \over {10}}} \right)}^2} + {{\left( {{\lambda \over {10}} - \lambda } \right)}^2}} \Rightarrow BP = \sqrt {{{9{\lambda ^2}} \over {100}} + {{81{\lambda ^2}} \over {100}}} = BP = \sqrt {{{90{\lambda ^2}} \over {100}}} Now, PA = \sqrt {{{\left( {{\lambda \over 3} - {{3\lambda } \over {10}}} \right)}^2} + {{\left( {0 - {\lambda \over {10}}} \right)}^2}} \Rightarrow$$\,\,\,$ PA = $\sqrt {{{{\lambda ^2}} \over {900}} + {{{\lambda ^2}} \over {100}}} \Rightarrow PA$ = $\sqrt {{{10{\lambda ^2}} \over {900}}}$

Therefore BP : PA = 9 : 1

3

### JEE Main 2019 (Online) 9th January Morning Slot

Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements is true ?
A
The lines are not concurrent
B
The lines are concurrent at the point $\left( {{3 \over 4},{1 \over 2}} \right)$
C
The lines are all parallel
D
Each line passes through the origin

## Explanation

Equation of lines;

px + qy + r = 0   . . . . . (1)

Also given

3p + 2q + 4r = 0   . . . . . . (2)

divide equation (2) by 4, we get

${3 \over 4}P + {2 \over 4}q + r = 0$   . . . . (3)

By comparing (1) and (3) we get,

x = ${3 \over 4}$ and y = ${2 \over 4}$ = ${1 \over 2}$

For any value of p,q and r, the equation of set of lines will pan through $\left( {{3 \over 4},{1 \over 2}} \right)$
4

### JEE Main 2019 (Online) 9th January Evening Slot

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 :
A
122y $-$ 26x $-$ 1675 = 0
B
122y + 26x + 1675 = 0
C
26x + 61y + 1675 = 0
D
26x $-$ 122y $-$ 1675 = 0

## Explanation

4x + 5y $-$ 20 = 0       . . .(1)

3x $-$ 2y + 6 = 0       . . . (2)

orthocentre is (1, 1)

line perpendicular to 4x + 5y $-$ 20 = 0

and passes through (1, 1) is

(y $-$ 1) = ${5 \over 4}$(x $-$ 1)

$\Rightarrow$  5x $-$ 4y = 1       . . .(3)

and line $\bot$ to 3x $-$ 2y + 6 = 0

and passes through (1, 1)

y $-$ 1 = $-$ ${2 \over 3}$ (x $-$ 1)

$\Rightarrow$  2x + 3y = 5       . . .(4)

Solving (1) and (4) we get C$\left( {{{35} \over 2}, - 10} \right)$

Solving (2) and (3) we get A $\left( { - 13,{{ - 33} \over 2}} \right)$

Side BC is y + 10 = ${{{{ - 33} \over 2} + 10} \over { - 13 - {{35} \over 2}}}\left( {x - {{35} \over 2}} \right)$

$\Rightarrow$  y + 10 = ${{13} \over {61}}\left( {x - {{35} \over 2}} \right)$

$\Rightarrow$  26x $-$ 122y $-$ 1675 = 0

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