Let the area of the triangle formed by a straight Line L: x + by + c = 0$L: x + by + c = 0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L$L$ makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, then the value of b^2 + c^2$b^2 + c^2$ is:
A.90
B.93
C.97
D.83
Solution & Explanation
### Related Formula
textNormal form of a straight line: x cos alpha + y sin alpha = p$$\text{Normal form of a straight line: } x \cos \alpha + y \sin \alpha = p$$textArea of right-angled triangle formed with axes: A = frac12 left| x_textintercept cdot y_textintercept right|$$\text{Area of right-angled triangle formed with axes: } A = \frac{1}{2} \left| x_{\text{intercept}} \cdot y_{\text{intercept}} \right|$$
### Core Logic
We write down the normal equation of the straight line using the given polar normal angle alpha = 45^circ$\alpha = 45^\circ$, find its intercept coordinates, and use the area constraint to solve for the coefficients.
### Step 1: Write down normal form
The perpendicular drawn from the origin makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, so alpha = 45^circ$\alpha = 45^\circ$. The line equation is:
x cos 45^circ + y sin 45^circ = p implies fracxsqrt2 + fracysqrt2 = p$$x \cos 45^\circ + y \sin 45^\circ = p \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = p$$x + y = psqrt2 implies x + y - psqrt2 = 0$$x + y = p\sqrt{2} \implies x + y - p\sqrt{2} = 0$$
Comparing this with the given format x + by + c = 0$x + by + c = 0$, we find:
b = 1 quad textand quad c = -psqrt2$$b = 1 \quad \text{and} \quad c = -p\sqrt{2}$$
### Step 2: Solve for the parameters using the area constraint
The line equation is x + y = psqrt2$x + y = p\sqrt{2}$. The intercepts are:
- x_textintercept = psqrt2$x_{\text{intercept}} = p\sqrt{2}$
- y_textintercept = psqrt2$y_{\text{intercept}} = p\sqrt{2}$
The area of the right-angled triangle formed with the axes is:
textArea = frac12 left| psqrt2 cdot psqrt2 right| = p^2$$\text{Area} = \frac{1}{2} \left| p\sqrt{2} \cdot p\sqrt{2} \right| = p^2$$
Since the area is given as 48 square units:
p^2 = 48$p^2 = 48$
### Step 3: Calculate the requested value
We have:
b^2 = 1^2 = 1$b^2 = 1^2 = 1$c^2 = (-psqrt2)^2 = 2p^2 = 2(48) = 96$$c^2 = (-p\sqrt{2})^2 = 2p^2 = 2(48) = 96$$
Therefore, we find:
b^2 + c^2 = 1 + 96 = 97$$b^2 + c^2 = 1 + 96 = 97$$
### Pattern Recognition
Normal equation coupling: Normal equations of lines xcosalpha+ysinalpha = p$x\cos\alpha+y\sin\alpha = p$ are extremely powerful when normal angles are specified. For alpha=45^circ$\alpha=45^\circ$, the coordinate intercepts are identical, making the area relation A=p^2$A=p^2$ exceptionally simple.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Keywords:#normal form straight line intercept#JEE Main 2025 Evening Q69#triangle area coordinate axis#analytical straight lines coefficients
More Straight Lines Previous-Year Questions — Page 3
Q51jee_main_2025_07_april_eveningOrthocentre of a Triangle
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
### Related Formula
The product of slopes of two mutually perpendicular lines is always equal to -1$-1$:
m_1 cdot m_2 = -1$$m_1 \cdot m_2 = -1$$
### Core Logic
Let the vertices of the triangle be P$P$, Q$Q$, and R$R$. The lines given are:
1) y = x + 1$y = x + 1$
2) y = 4x - 8$y = 4x - 8$
3) y = mx + c$y = mx + c$
Solving lines y = x + 1$y = x + 1$ and y = 4x - 8$y = 4x - 8$ gives the vertex P(3, 4)$P(3, 4)$.
The orthocentre is given as H(3, -1)$H(3, -1)$. Notice that the x$x$-coordinate of P$P$ and H$H$ are identical (x = 3$x = 3$). This implies that the altitude from vertex P$P$ to the base line y = mx + c$y = mx + c$ is a vertical line along x = 3$x = 3$.
Orthocentre of a Triangle diagram for Q51 - JEE Main 2025 Evening
### Step 1: Determine the Slopes
Since the altitude from P$P$ is vertical, the side opposite to it (which lies on y = mx + c$y = mx + c$) must be a horizontal line.
Therefore, the slope of the line y = mx + c$y = mx + c$ must be zero:
m = 0$m = 0$
### Step 2: Solve for c
Let's find point Q$Q$ by intersecting y = x + 1$y = x + 1$ and y = mx + c$y = mx + c$. Since m = 0$m = 0$, y = c$y = c$, we get Q(c-1, c)$Q(c-1, c)$.
Using the property that the line segment connecting Q$Q$ to the opposite side's altitude is perpendicular to line PR$PR$ (y = 4x - 8$y = 4x - 8$):
textSlope of QH cdot textSlope of PR = -1$$\text{Slope of } QH \cdot \text{Slope of } PR = -1$$frac-1 - c3 - (c - 1) cdot 4 = -1$$\frac{-1 - c}{3 - (c - 1)} \cdot 4 = -1$$frac-4(c + 1)4 - c = -1 implies 4c + 4 = 4 - c implies 5c = 0 implies c = 0$$\frac{-4(c + 1)}{4 - c} = -1 \implies 4c + 4 = 4 - c \implies 5c = 0 \implies c = 0$$
### Step 3: Evaluate m - c
Substituting the values of m$m$ and c$c$:
m - c = 0 - 0 = 0$$m - c = 0 - 0 = 0$$
### Pattern Recognition
When the x$x$-coordinate of a vertex matches the x$x$-coordinate of the orthocentre, the altitude is vertical, forcing the opposite base to be purely horizontal (m=0$m=0$). This observation cuts down calculation time completely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q69jee_main_2025_24_jan_morningConcurrency of Straight Lines
Let the lines 3x - 4y - alpha = 0$3x - 4y - \alpha = 0$, 8x - 11y - 33 = 0$8x - 11y - 33 = 0$, and 2x - 3y + lambda = 0$2x - 3y + \lambda = 0$ be concurrent. If the image of the point (1, 2)$(1, 2)$ in the line 2x - 3y + lambda = 0$2x - 3y + \lambda = 0$ is left(frac5713,frac-4013
ight)$\left(\frac{57}{13},\frac{-40}{13}
ight)$ , then |alpha lambda|$|\alpha \lambda|$ 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$M$ between point P(1, 2)$P(1, 2)$ and its given reflection image Qleft(frac5713, frac-4013right)$Q\left(\frac{57}{13}, \frac{-40}{13}\right)$:
M = left( frac1 + frac57132, \, frac2 - frac40132 right) = left( frac7026, \, frac-1426 right) = left( frac3513, \, frac-713 right)$$M = \left( \frac{1 + \frac{57}{13}}{2}, \, \frac{2 - \frac{40}{13}}{2} \right) = \left( \frac{70}{26}, \, \frac{-14}{26} \right) = \left( \frac{35}{13}, \, \frac{-7}{13} \right)$$
Since M$M$ lies on the reflecting line 2x - 3y + lambda = 0$2x - 3y + \lambda = 0$:
2left(frac3513right) - 3left(frac-713right) + lambda = 0$$2\left(\frac{35}{13}\right) - 3\left(\frac{-7}{13}\right) + \lambda = 0$$frac7013 + frac2113 + lambda = 0 implies frac9113 + lambda = 0 implies 7 + lambda = 0 implies lambda = -7$$\frac{70}{13} + \frac{21}{13} + \lambda = 0 \implies \frac{91}{13} + \lambda = 0 \implies 7 + \lambda = 0 \implies \lambda = -7$$
### 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$$\left| \begin{matrix} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & -3 & -7 \end{matrix} \right| = 0$$
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$$3\left[ (-11)(-7) - (-33)(-3) \right] - (-4)\left[ (8)(-7) - (-33)(2) \right] - \alpha \left[ (8)(-3) - (-11)(2) \right] = 0$$3[77 - 99] + 4[-56 + 66] - alpha[-24 + 22] = 0$$3[77 - 99] + 4[-56 + 66] - \alpha[-24 + 22] = 0$$3[-22] + 4[10] - alpha[-2] = 0$$3[-22] + 4[10] - \alpha[-2] = 0$$-66 + 40 + 2alpha = 0 implies 2alpha = 26 implies alpha = 13$$-66 + 40 + 2\alpha = 0 \implies 2\alpha = 26 \implies \alpha = 13$$
### Step 2: Compute Final Product Target
Multiply the absolute values of the determined parameters together:
|alpha lambda| = |13 cdot (-7)| = |-91| = 91$$|\alpha \lambda| = |13 \cdot (-7)| = |-91| = 91$$
### 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
Q67jee_main_2025_28_jan_eveningAngle Between Lines
Two equal sides of an isosceles triangle are along -x+2y=4$-x+2y=4$ and x+y=4.$x+y=4.$ 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$\theta$ between two lines with slopes m_1$m_1$ and m_2$m_2$:
tantheta = left| fracm_1 - m_21 + m_1 m_2 right|$$\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$
### Core Logic
Given lines for the equal sides:
1) -x + 2y = 4 implies y = frac12x + 2 implies m_1 = frac12$-x + 2y = 4 \implies y = \frac{1}{2}x + 2 \implies m_1 = \frac{1}{2}$
2) x + y = 4 implies y = -x + 4 implies m_2 = -1$x + y = 4 \implies y = -x + 4 \implies m_2 = -1$
In an isosceles triangle, the third side makes equal angles theta$\theta$ with both equal sides. Let the slope of the third side be m$m$:
left| fracm - 1/21 + m/2 right| = left| fracm - (-1)1 + m(-1) right|$$\left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{m - (-1)}{1 + m(-1)} \right|$$left| frac2m - 12 + m right| = left| fracm + 11 - m right|$$\left| \frac{2m - 1}{2 + m} \right| = \left| \frac{m + 1}{1 - m} \right|$$
### Step 1: Solve the Slope Equation
Case 1 (Same sign):
frac2m - 12 + m = fracm + 11 - m$$\frac{2m - 1}{2 + m} = \frac{m + 1}{1 - m}$$(2m - 1)(1 - m) = (m + 1)(2 + m)$$(2m - 1)(1 - m) = (m + 1)(2 + m)$$2m - 2m^2 - 1 + m = m^2 + 3m + 2$$2m - 2m^2 - 1 + m = m^2 + 3m + 2$$-2m^2 + 3m - 1 = m^2 + 3m + 2$$-2m^2 + 3m - 1 = m^2 + 3m + 2$$3m^2 + 3 = 0 implies m^2 = -1 quad (textNo real roots)$$3m^2 + 3 = 0 \implies m^2 = -1 \quad (\text{No real roots})$$
Case 2 (Opposite sign):
frac2m - 12 + m = -fracm + 11 - m = fracm + 1m - 1$$\frac{2m - 1}{2 + m} = -\frac{m + 1}{1 - m} = \frac{m + 1}{m - 1}$$(2m - 1)(m - 1) = (2 + m)(m + 1)$$(2m - 1)(m - 1) = (2 + m)(m + 1)$$2m^2 - 3m + 1 = m^2 + 3m + 2$$2m^2 - 3m + 1 = m^2 + 3m + 2$$m^2 - 6m - 1 = 0$$m^2 - 6m - 1 = 0$$
### Step 2: Sum of Roots
The quadratic equation for m$m$ is m^2 - 6m - 1 = 0$m^2 - 6m - 1 = 0$.
The sum of possible distinct values of m$m$ is given by the sum of roots of this quadratic:
textSum of roots = -frac-61 = 6$$\text{Sum of roots} = -\frac{-6}{1} = 6$$
### 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$m^2 - 6m - 1 = 0$ gives the final answer instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q54jee_main_2025_29_jan_morningCentroid and Image of a Point
Let ABC be a triangle formed by the lines 7mathrmx - 6mathrmy + 3 = 0, mathrmx + 2mathrmy - 31 = 0$7\mathrm{x} - 6\mathrm{y} + 3 = 0, \mathrm{x} + 2\mathrm{y} - 31 = 0$ and 9mathrmx - 2mathrmy - 19 = 0$9\mathrm{x} - 2\mathrm{y} - 19 = 0$ . Let the point (mathrmh,mathrmk)$(\mathrm{h},\mathrm{k})$ be the image of the centroid of Delta ABC$\Delta ABC$ in the line 3mathrmx + 6mathrmy - 53 = 0$3\mathrm{x} + 6\mathrm{y} - 53 = 0$ . 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)$$\text{Centroid } G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$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$$\text{Image of } (x_1, y_1) \text{ in line } ax+by+c=0: \frac{x-x_1}{a} = \frac{y-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$$
### Core Logic
First, find the vertices A, B, C$A, B, C$ by solving the lines pairwise.
Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and x + 2y - 31 = 0$x + 2y - 31 = 0$ gives A(9,11)$A(9,11)$.
Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives B(3,4)$B(3,4)$.
Solving x + 2y - 31 = 0$x + 2y - 31 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives C(5,13)$C(5,13)$.
Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 1: Determine the Centroid
G = left(frac9 + 3 + 53, frac11 + 4 + 133right) = left(frac173, frac283right)$$G = \left(\frac{9 + 3 + 5}{3}, \frac{11 + 4 + 13}{3}\right) = \left(\frac{17}{3}, \frac{28}{3}\right)$$
### Step 2: Find the Image (h, k)
Using the line 3x + 6y - 53 = 0$3x + 6y - 53 = 0$:
frach - frac1733 = frack - frac2836 = -2 frac3left(frac173right) + 6left(frac283right) - 533^2 + 6^2$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -2 \frac{3\left(\frac{17}{3}\right) + 6\left(\frac{28}{3}\right) - 53}{3^2 + 6^2}$$frach - frac1733 = frack - frac2836 = -2 frac17 + 56 - 5345 = -2 frac2045 = -frac89$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -2 \frac{17 + 56 - 53}{45} = -2 \frac{20}{45} = -\frac{8}{9}$$
Solving for h$h$ and k$k$ yields:
h = 3, quad k = 4$$h = 3, \quad k = 4$$Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 3: Compute final algebraic target
h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37$$h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37$$
### Pattern Recognition
Instead of solving fractions endlessly, substitute potential integer coordinates early into the slope relationship (k - y_G)/(h - x_G) = -1/m$(k - y_G)/(h - x_G) = -1/m$ to accelerate competitive solving time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q11jee_main_2024_29_january_eveningDistance of a Point From a Line
The distance of the point (2, 3) from the line 2x - 3y + 28 = 0$2x - 3y + 28 = 0$, measured parallel to the line sqrt3 x - y + 1 = 0$sqrt{3} x - y + 1 = 0$, is equal to
A.4sqrt2$4\sqrt{2}$
B.6sqrt3$6\sqrt{3}$
C.3 + 4sqrt2$3 + 4\sqrt{2}$
D.4 + 6sqrt3$4 + 6\sqrt{3}$
Solution
### Related Formula
x = x_1 + r cos theta, quad y = y_1 + r sin theta$$x = x_1 + r \cos \theta, \quad y = y_1 + r \sin \theta$$
### Core Logic
The line is measured parallel to sqrt3x - y + 1 = 0$\sqrt{3}x - y + 1 = 0$, which has a slope tan theta = sqrt3 implies theta = 60^circ$\tan \theta = \sqrt{3} \implies \theta = 60^\circ$.
Thus, cos theta = frac12$\cos \theta = \frac{1}{2}$ and \sin \theta = \frac{\sqrt{3}}{2}.
Writing any point $.
Writing any point $P along this direction passing through $ along this direction passing through $(2,3) in parametric coordinates:
$ in parametric coordinates:
$P = left(2 + r cos 60^circ, 3 + r sin 60^circright) = left(2 + fracr2, 3 + fracsqrt3r2right)$P = \left(2 + r \cos 60^\circ, 3 + r \sin 60^\circ\right) = \left(2 + \frac{r}{2}, 3 + \frac{\sqrt{3}r}{2}\right)$
### Step 1: Finding Intersection Point
Since $
### Step 1: Finding Intersection Point
Since $P must lie on the given line $ must lie on the given line $2x - 3y + 28 = 0:
$:
$2left(2 + fracr2right) - 3left(3 + fracsqrt3r2right) + 28 = 0$2\left(2 + \frac{r}{2}\right) - 3\left(3 + \frac{\sqrt{3}r}{2}\right) + 28 = 0$$
$4 + r - 9 - frac3sqrt3r2 + 28 = 0$4 + r - 9 - \frac{3\sqrt{3}r}{2} + 28 = 0$$
$23 + rleft(1 - frac3sqrt32right) = 0$23 + r\left(1 - \frac{3\sqrt{3}}{2}\right) = 0$$
$rleft(frac3sqrt3 - 22right) = 23 implies r = frac463sqrt3 - 2$r\left(\frac{3\sqrt{3} - 2}{2}\right) = 23 \implies r = \frac{46}{3\sqrt{3} - 2}$
Rationalizing the denominator:
$
Rationalizing the denominator:
$r = frac46(3sqrt3 + 2)(3sqrt3)^2 - 2^2 = frac46(3sqrt3 + 2)27 - 4 = frac46(3sqrt3 + 2)23 = 2(3sqrt3 + 2) = 4 + 6sqrt3$r = \frac{46(3\sqrt{3} + 2)}{(3\sqrt{3})^2 - 2^2} = \frac{46(3\sqrt{3} + 2)}{27 - 4} = \frac{46(3\sqrt{3} + 2)}{23} = 2(3\sqrt{3} + 2) = 4 + 6\sqrt{3}$$
### Pattern Recognition
Distance measured parallel to a given direction is always resolved most efficiently using parametric equations of lines rather than perpendicular metrics.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
More Straight Lines Questions — jee_main_2025_02_april_evening
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