1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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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