1
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
2
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to -
A
$${7 \over 4}$$
B
$${5 \over 4}$$
C
$${3 \over 4}$$
D
$${3 \over 2}$$

Explanation

r = 1 is obviously true.

Let 0 < r < 1

$$ \Rightarrow $$  r + r2 > 1

$$ \Rightarrow $$  r2 + r $$-$$ 1 > 0

$$\left( {r - {{ - 1 - \sqrt 5 } \over 2}} \right)\left( {r - \left( {{{ - 1 + \sqrt 5 } \over 2}} \right)} \right)$$

$$ \Rightarrow r - {{ - 1 - \sqrt 5 } \over 2}$$  or  $$r > {{ - 1 + \sqrt 5 } \over 2}$$

$$r \in \left( {{{\sqrt 5 - 1} \over 2},1} \right)$$

$${{\sqrt 5 - 1} \over 2} < r < 1$$

When r > 1

$$ \Rightarrow {{\sqrt 5 + 1} \over 2} > {1 \over r} > 1$$

$$ \Rightarrow r \in \left( {{{\sqrt 5 - 1} \over 2},{{\sqrt 5 + 1} \over 2}} \right)$$

Now check options
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R (3, – 2) are fixed points, then the locus of the centroid of $$\Delta $$PQR is a line -
A
parallel to y-axis
B
with slope $${2 \over 3}$$
C
parallel to x-axis
D
with slope $${3 \over 2}$$

Explanation

Let the centroid of $$\Delta $$PQR is (h, k) & P is ($$\alpha $$, $$\beta $$), then

$${{\alpha + 1 + 3} \over 3} = h\,$$   and   $${{\beta + 4 - 2} \over 3} = k$$

$$\alpha = \left( {3h - 4} \right)$$   $$\beta = \left( {3k - 4} \right)$$

Point P($$\alpha $$, $$\beta $$) lies on the line 2x $$-$$ 3y + 4 = 0

$$ \therefore $$  2(3h $$-$$ 4) $$-$$ 3 (3k $$-$$ 2) + 4 = 0

$$ \Rightarrow $$  locus is 6x $$-$$ 9y + 2 = 0
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 10th January Morning Slot

If the line 3x + 4y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is -
A
(3, 4)
B
(2, 2)
C
(4, 4)
D
(4, 3)

Explanation



$$\left| {{{3r + 4r - 24} \over 5}} \right| = r$$

$$7r - 24 = \pm 5r$$

$$2r = 24$$  or  $$12r + 24$$

$$r = 14,\,\,\,r = 2$$

then incentre is $$(2,2)$$

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