Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 unit

Q: Why use parametric form? Can't I just use y - 3 = m(x - 2) and the perpendicular distance formula?

You absolutely can, but the perpendicular distance formula gives you the distance from the point to the line, not the distance along the line to the intersection. The parametric form is specifically designed to handle 'distance along a line' problems elegantly, saving you from dealing with messy radical equations.

Q: Why do we get a quadratic equation for m?

Because there are two distinct lines passing through P that intersect the target line at a distance of 4 units. One line tilts one way, the other tilts the other way. The quadratic captures both possibilities, which is why you see the plus-minus sign in the final answer.

Q: My answer is \frac{-4 \pm \sqrt{7}}{3}, but the options look different. Did I go wrong?

You are likely correct! In JEE Advanced, options are often presented in rationalized or alternative forms. Try rationalizing the denominator of the given options to see if they match your result. It is a test of your algebraic manipulation skills, not a sign of failure.

Question

Slope of a line passing through $P(2, 3)$ and intersecting the line, $x + y = 7$ at a distance of 4 units from $P$, is

Options:

(A) $\frac{\sqrt{5}-1}{\sqrt{5}+1}$

(B) $\frac{1-\sqrt{5}}{1+\sqrt{5}}$

(C) $\frac{1-\sqrt{7}}{1+\sqrt{7}}$

(D) $\frac{\sqrt{7}-1}{\sqrt{7}+1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are embarking on a journey to understand how lines interact in space.

Imagine you are standing at the point $P(2, 3)$ on a vast, flat plane. In front of you lies a straight road defined by the equation $x + y = 7$.

You want to walk from your current position $P$ to this road, but you have a very specific constraint: you must walk exactly $4$ units to reach it. This is the core of our problem.

If you visualize this, you will see that the set of all points exactly $4$ units away from $P$ forms a circle centered at $P$ with a radius of $4$. The points where this circle intersects our target line are the destinations of your journey.

### The Power of Parametric Coordinates

Now, how do we translate this visual intuition into the language of algebra? We could use the standard slope-intercept form, but that often leads to messy square roots and complex calculations.

Instead, we use the most elegant tool in our arsenal: the parametric form of a line. If a line passes through a point $(x_0, y_0)$ and makes an angle $	heta$ with the positive x-axis, any point on that line at a distance $r$ from the origin can be described as $(x_0 + r \cos 	heta, y_0 + r \sin 	heta)$.

In our case, $P$ is $(2, 3)$ and $r$ is $4$. So, any point on our path is simply $(2 + 4 \cos 	heta, 3 + 4 \sin 	heta)$. This is like having a GPS coordinate for every point on your path, dependent only on the angle $	heta$ at which you choose to walk.

### The Algebraic Dance

Since the point where you hit the road must lie on the line $x + y = 7$, we can substitute our parametric coordinates directly into this equation.

Substituting $x = 2 + 4 \cos 	heta$ and $y = 3 + 4 \sin 	heta$ into $x + y = 7$, we get:

$$ (2 + 4 \cos 	heta) + (3 + 4 \sin 	heta) = 7 $$

Let us simplify this step-by-step. Combining the constants $2$ and $3$ gives us $5$. Subtracting $5$ from both sides leaves us with $4(\cos 	heta + \sin 	heta) = 2$.

Dividing by $4$, we arrive at the beautiful, simplified relation:

$$ \cos 	heta + \sin 	heta = \frac{1}{2} $$

### The Trigonometric Bridge

We are almost there, but we need the slope $m = 	an 	heta$. How do we get from $\cos 	heta + \sin 	heta$ to $	an 	heta$? We square both sides!

Squaring $\cos 	heta + \sin 	heta = \frac{1}{2}$ gives us:

$$ \cos^2 	heta + \sin^2 	heta + 2 \sin 	heta \cos 	heta = \frac{1}{4} $$

Using the fundamental identity $\cos^2 	heta + \sin^2 	heta = 1$ and the double angle identity $2 \sin 	heta \cos 	heta = \sin 2	heta$, we get $1 + \sin 2	heta = \frac{1}{4}$, which simplifies to $\sin 2	heta = -\frac{3}{4}$.

Now, we use the classic JEE identity that connects the double angle to the slope:

$$ \sin 2	heta = \frac{2m}{1 + m^2} $$

Substituting our value, we get:

$$ \frac{2m}{1 + m^2} = -\frac{3}{4} $$

### The Final Resolution

Cross-multiplying gives us $8m = -3(1 + m^2)$, which rearranges into the quadratic equation:

$$ 3m^2 + 8m + 3 = 0 $$

Solving this using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find:

$$ m = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-4 \pm \sqrt{7}}{3} $$

You might look at the options and feel a moment of doubt because they don't look exactly like this. But remember, in mathematics, form is often a matter of perspective.

By rationalizing the denominator of the options, you will find that they are identical to our result. You have successfully navigated the geometry, the algebra, and the trigonometry. **The final slopes are $m = \frac{-4 \pm \sqrt{7}}{3}$.**

Animated Solution for Mathematics - Straight Lines: Slope of a line passing through $P (2, 3)$ and intersecting the line, $x + y = 7$ at a distance of 4 units from $P$ , is

Select Answer:

Visualized Solution

Visualizing the Setup

The Distance Constraint

Parametric Coordinates

Intersection Condition

Simplifying the Equation

Squaring to Find $sin 2 θ$

Relating to Slope $m$

Forming the Equation in $m$

Solving for $m$

Matching the Options

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Power of Parametric Coordinates

The Algebraic Dance

The Trigonometric Bridge

The Final Resolution

Similar Questions

If a straight line passing through the point $P (- 3, 4)$ is such that its intercepted portion between the coordinate axes is bisected at $P$ , then its equation is :

The distance of the point $(2, 3)$ from the line $2 x - 3 y + 28 = 0$ , measured parallel to the line $3 x - y + 1 = 0$ , is equal to

A straight line through the point $A (3, 4)$ is such that its intercept between the axes is bisected at $A$ . Its equation is

The equation of the straight line passing through the point $(4, 3)$ and making intercepts on the co-ordinate axes whose sum is $- 1$ is

Let the angles made with the positive $x$ -axis by two straight lines drawn from the point $P (2, 3)$ and meeting the line $x + y = 6$ at a distance $\frac{2}{3}$ from the point P be $θ_{1}$ and $θ_{2}$ . Then the value of $(θ_{1} + θ_{2})$ is:

Suppose that the points (h,k), (1,2) and (-3,4) lie on the line L1. If a line L2 passing through the points (h,k) and (4,3) is perpendicular to L1, then k/h equals :

Let $P S$ be the median of the triangle with vertices $P (2, 2), Q (6, - 1)$ and $R (7, 3)$ . The equation of the line passing through $(1, - 1)$ and parallel to $P S$ is

A line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a triangle $O P Q$ , where $O$ is the origin. If the area of the triangle $O P Q$ is least, then the slope of the line $P Q$ is

Let $A B C$ be an isosceles triangle in which $A$ is at $(- 1, 0)$ , $∠ A = \frac{2 π}{3}$ , $A B = A C$ and $B$ is on the positive $x$ -axis. If $B C = 43$ and the line $B C$ intersects the line $y = x + 3$ at $(α, β)$ , then $\frac{β ^{4}}{α ^{2}}$ is :

The straight lines $l_{1}$ and $l_{2}$ pass through the origin and trisect the line segment of the line $L : 9 x + 5 y = 45$ between the axes. If $m_{1}$ and $m_{2}$ are the slopes of the lines $l_{1}$ and $l_{2}$ , then the point of intersection of the line $y = (m_{1} + m_{2}) x$ with $L$ lies on

Animated Solution for Mathematics - Straight Lines: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Slope of a line passing through P(2,3) and intersecting the line, x+y=7 at a distance of 4 units from P, is