If the orthocentre of the triangle formed by the lines $y = x + 1$ $y = x + 1$, $y = 4x - 8$ $y = 4x - 8$ and $y = mx + c$ $y = mx + c$ is at $(3, -1)$ $(3, -1)$, then $m - c$ $m - c$ is:

A. $0$ $0$
B. $-2$ $-2$
C. $4$ $4$
D. $2$ $2$

Solution & Explanation

### Related Formula The product of slopes of two mutually perpendicular lines is always equal to

-1

m_1 cdot m_2 = -1

### Core Logic Let the vertices of the triangle be

P

Q

, and

R

. The lines given are: 1)

y = x + 1

y = 4x - 8

y = mx + c

Solving lines

y = x + 1

and

y = 4x - 8

gives the vertex

P(3, 4)

. The orthocentre is given as

H(3, -1)

. Notice that the

x

-coordinate of

P

and

H

are identical (

x = 3

). This implies that the altitude from vertex

P

to the base line

y = mx + c

is a vertical line along

x = 3

Orthocentre of a Triangle diagram for Q51 - JEE Main 2025 Evening

### Step 1: Determine the Slopes Since the altitude from

P

is vertical, the side opposite to it (which lies on

y = mx + c

) must be a horizontal line. Therefore, the slope of the line

y = mx + c

must be zero:

m = 0

### Step 2: Solve for c Let's find point

Q

by intersecting

y = x + 1

and

y = mx + c

. Since

m = 0

y = c

, we get

Q(c-1, c)

. Using the property that the line segment connecting

Q

to the opposite side's altitude is perpendicular to line

PR

(

y = 4x - 8

textSlope of QH cdot textSlope of PR = -1

frac-1 - c3 - (c - 1) cdot 4 = -1

frac-4(c + 1)4 - c = -1 implies 4c + 4 = 4 - c implies 5c = 0 implies c = 0

### Step 3: Evaluate m - c Substituting the values of

m

and

c

m - c = 0 - 0 = 0

### Pattern Recognition When the

x

-coordinate of a vertex matches the

x

-coordinate of the orthocentre, the altitude is vertical, forcing the opposite base to be purely horizontal (

m=0

). This observation cuts down calculation time completely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

Reference Study Guides

Straight Lines

More Straight Lines Previous-Year Questions — Page 3

Q69 2025 Concurrency of Straight Lines

Let the lines

3x - 4y - alpha = 0

8x - 11y - 33 = 0

, and

2x - 3y + lambda = 0

be concurrent. If the image of the point

(1, 2)

in the line

2x - 3y + lambda = 0

left(frac5713,frac-4013 ight)

, then

|alpha lambda|

is equal to :

A. $84$ $84$
B. $91$ $91$
C. $113$ $113$
D. $101$ $101$

Solution

### Related Formula The midpoint between a point and its reflection image must lie exactly on the line mirror equation. ### Core Logic Find the midpoint

M

between point

P(1, 2)

and its given reflection image

Qleft(frac5713, frac-4013right)

M = left( frac1 + frac57132, \, frac2 - frac40132 right) = left( frac7026, \, frac-1426 right) = left( frac3513, \, frac-713 right)

Since

M

lies on the reflecting line

2x - 3y + lambda = 0

2left(frac3513right) - 3left(frac-713right) + lambda = 0

frac7013 + frac2113 + lambda = 0 implies frac9113 + lambda = 0 implies 7 + lambda = 0 implies lambda = -7

### Step 1: Apply Concurrency Determinant For three straight lines to intersect at a single concurrent point, the determinant of their linear coefficients must equal zero:

left| beginmatrix 3 & -4 & -alpha \\ 8 & -11 & -33 \\ 2 & -3 & -7 endmatrix right| = 0

Expand the determinant along the first row:

3left[ (-11)(-7) - (-33)(-3) right] - (-4)left[ (8)(-7) - (-33)(2) right] - alpha left[ (8)(-3) - (-11)(2) right] = 0

3[77 - 99] + 4[-56 + 66] - alpha[-24 + 22] = 0

3[-22] + 4[10] - alpha[-2] = 0

-66 + 40 + 2alpha = 0 implies 2alpha = 26 implies alpha = 13

### Step 2: Compute Final Product Target Multiply the absolute values of the determined parameters together:

|alpha lambda| = |13 cdot (-7)| = |-91| = 91

### Pattern Recognition Using the midpoint property to evaluate unknown line parameters from reflection images is often much faster than using full distance formulas. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

Q67 2025 Angle Between Lines

Two equal sides of an isosceles triangle are along

-x+2y=4

and

x+y=4.

If m is the slope of its third side, then the sum, of all possible distinct values of m, is:

A. $-6$ $-6$
B. $12$ $12$
C. $6$ $6$
D. $-2sqrt10$ $-2\sqrt{10}$

Solution

### Related Formula Angle

theta

between two lines with slopes

m_1

and

m_2

tantheta = left| fracm_1 - m_21 + m_1 m_2 right|

### Core Logic Given lines for the equal sides: 1)

-x + 2y = 4 implies y = frac12x + 2 implies m_1 = frac12

x + y = 4 implies y = -x + 4 implies m_2 = -1

In an isosceles triangle, the third side makes equal angles

theta

with both equal sides. Let the slope of the third side be

m

left| fracm - 1/21 + m/2 right| = left| fracm - (-1)1 + m(-1) right|

left| frac2m - 12 + m right| = left| fracm + 11 - m right|

### Step 1: Solve the Slope Equation Case 1 (Same sign):

frac2m - 12 + m = fracm + 11 - m

(2m - 1)(1 - m) = (m + 1)(2 + m)

2m - 2m^2 - 1 + m = m^2 + 3m + 2

-2m^2 + 3m - 1 = m^2 + 3m + 2

3m^2 + 3 = 0 implies m^2 = -1 quad (textNo real roots)

Case 2 (Opposite sign):

frac2m - 12 + m = -fracm + 11 - m = fracm + 1m - 1

(2m - 1)(m - 1) = (2 + m)(m + 1)

2m^2 - 3m + 1 = m^2 + 3m + 2

m^2 - 6m - 1 = 0

### Step 2: Sum of Roots The quadratic equation for

m

m^2 - 6m - 1 = 0

. The sum of possible distinct values of

m

is given by the sum of roots of this quadratic:

textSum of roots = -frac-61 = 6

### Pattern Recognition Instead of solving for the explicit values of the slopes (which involve radicals), using Vieta's relations directly on the quadratic equation

m^2 - 6m - 1 = 0

gives the final answer instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

Q54 2025 Centroid and Image of a Point

Let ABC be a triangle formed by the lines

7mathrmx - 6mathrmy + 3 = 0, mathrmx + 2mathrmy - 31 = 0

and

9mathrmx - 2mathrmy - 19 = 0

. Let the point

(mathrmh,mathrmk)

be the image of the centroid of

Delta ABC

in the line

3mathrmx + 6mathrmy - 53 = 0

. Then \mathrm{h}^2 + \mathrm{k}^2 + \mathrm{hk} is equal to

A. 37
B. 47
C. 40
D. 36

Solution

### Related Formula

textCentroid G = left(fracx_1+x_2+x_33, fracy_1+y_2+y_33right)

textImage of (x_1, y_1) text in line ax+by+c=0: fracx-x_1a = fracy-y_1b = -2fracax_1+by_1+ca^2+b^2

### Core Logic First, find the vertices

A, B, C

by solving the lines pairwise. Solving

7x - 6y + 3 = 0

and

x + 2y - 31 = 0

gives

A(9,11)

. Solving

7x - 6y + 3 = 0

and

9x - 2y - 19 = 0

gives

B(3,4)

. Solving

x + 2y - 31 = 0

and

9x - 2y - 19 = 0

gives

C(5,13)

Centroid diagram for Q54 - JEE Main 2025 Morning

### Step 1: Determine the Centroid

G = left(frac9 + 3 + 53, frac11 + 4 + 133right) = left(frac173, frac283right)

### Step 2: Find the Image (h, k) Using the line

3x + 6y - 53 = 0

frach - frac1733 = frack - frac2836 = -2 frac3left(frac173right) + 6left(frac283right) - 533^2 + 6^2

frach - frac1733 = frack - frac2836 = -2 frac17 + 56 - 5345 = -2 frac2045 = -frac89

Solving for

h

and

k

yields:

h = 3, quad k = 4

### Step 3: Compute final algebraic target

h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37

### Pattern Recognition Instead of solving fractions endlessly, substitute potential integer coordinates early into the slope relationship

(k - y_G)/(h - x_G) = -1/m

to accelerate competitive solving time. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

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If the orthocentre of the triangle formed by the lines $y = x + 1$ $y = x + 1$, $y = 4x - 8$ $y = 4x - 8$ and $y = mx + c$ $y = mx + c$ is at $(3, -1)$ $(3, -1)$, then $m - c$ $m - c$ is:

Solution & Explanation

Reference Study Guides

More Straight Lines Previous-Year Questions — Page 3

Solution

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If the orthocentre of the triangle formed by the lines y = x + 1$y = x + 1$, y = 4x - 8$y = 4x - 8$ and y = mx + c$y = mx + c$ is at (3, -1)$(3, -1)$, then m - c$m - c$ is:

Solution & Explanation

Reference Study Guides

More Straight Lines Previous-Year Questions — Page 3

Solution

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If the orthocentre of the triangle formed by the lines $y = x + 1$ $y = x + 1$, $y = 4x - 8$ $y = 4x - 8$ and $y = mx + c$ $y = mx + c$ is at $(3, -1)$ $(3, -1)$, then $m - c$ $m - c$ is: