If the straight line, 2x-3y+17 = 0 is perpendicular to the line passing through the points (7, 17) a

Q: Why is the product of slopes $-1$? Can't it be $1$?

Great question! If the product were $1$, the lines would be parallel (or identical). The $-1$ comes from the rotation of the coordinate system by $90^\circ$. When you rotate a line with slope $m$ by $90^\circ$, the new slope becomes $-1/m$.

Q: What if the line is vertical? Does the slope formula still work?

Excellent catch! If a line is vertical, its slope is undefined. In that case, the perpendicular line must be horizontal (slope $0$). Our formula $m_1 \cdot m_2 = -1$ works for all non-vertical/non-horizontal lines. Always check if your line is vertical before applying the formula!

Q: Can I use the vector dot product instead of slopes?

Absolutely! You are thinking like a physicist. If you treat the lines as vectors, their direction vectors must have a dot product of zero. It is a more robust method, especially in 3D geometry!

Question

If the straight line, $2x-3y+17 = 0$ is perpendicular to the line passing through the points $(7, 17)$ and $(15, \beta)$, then $\beta$ equals :-

Options:

(A) -5

(B) $-35/3$

(C) $35/3$

(D) 5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The essence of perpendicularity on a coordinate plane is defined by the relationship between the slopes of two lines. When two lines meet at a perfect $90^\circ$ angle, their slopes must satisfy a specific product constraint. We are tasked with finding the value of $\beta$ that forces this orthogonal condition between two given lines.

### Phase 1: The First Line

We begin with the equation $2x - 3y + 17 = 0$. To determine its slope, we transform it into the slope-intercept form, $y = mx + c$.

Shifting $3y$ to the right side, we obtain:
$$3y = 2x + 17$$

Dividing by $3$, we arrive at:
$$y = \frac{2}{3}x + \frac{17}{3}$$

Here, the coefficient of $x$ represents our slope, $m_1 = \frac{2}{3}$. This line climbs steadily, rising $2$ units for every $3$ units it moves to the right.

### Phase 2: The Second Line

The second line is defined by two points: $A(7, 17)$ and $B(15, \beta)$. The slope $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by the ratio of vertical change to horizontal change:
$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Applying this to our points, we calculate:
$$m_2 = \frac{\beta - 17}{15 - 7}$$

Simplifying the denominator, we find the expression for the second slope:
$$m_2 = \frac{\beta - 17}{8}$$

### Phase 3: The Perpendicularity Condition

The condition for two lines to be perpendicular is that the product of their slopes must be exactly $-1$. Mathematically, this is expressed as:
$$m_1 \cdot m_2 = -1$$

Substituting our derived values into this equation, we bridge the two phases:
$$\frac{2}{3} \cdot \left( \frac{\beta - 17}{8} \right) = -1$$

### Phase 4: The Final Calculation

To solve for $\beta$, we first multiply the fractions:
$$\frac{2(\beta - 17)}{24} = -1$$

Simplifying the fraction $\frac{2}{24}$ to $\frac{1}{12}$, we obtain:
$$\frac{\beta - 17}{12} = -1$$

Multiplying both sides by $12$ yields:
$$\beta - 17 = -12$$

Adding $17$ to both sides, we arrive at the final result:
$$\beta = 5$$

The value of **$\beta = 5$** is the unique solution that satisfies the geometric requirement of perpendicularity. This logic remains a fundamental tool for solving coordinate geometry problems in the JEE Advanced curriculum.

Animated Solution for Mathematics - Straight Lines: If the straight line, $2 x - 3 y + 17 = 0$ is perpendicular to the line passing through the points $(7, 17)$ and $(15, β)$ , then $β$ equals :-

Select Answer:

Visualized Solution

The Given Line

The Second Line

The Perpendicularity Condition

Slopes of Perpendicular Lines

Slope of the First Line

Calculating $m_{1}$

Slope of the Second Line

Substituting the Points

Simplifying $m_{2}$

Applying the Condition

Simplifying the Equation

Solving for $β$

Final Answer

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

If the line $\frac{2 - x}{3} = \frac{3 y - 2}{4 λ + 1} = 4 - z$ makes a right angle with the line $\frac{x + 3}{3 μ} = \frac{1 - 2 y}{6} = \frac{5 - z}{7}$ , then $4 λ + 9 μ$ is equal to :

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to

Straight lines $3 x + 4 y = 5$ and $4 x - 3 y = 15$ intersect at the point $A$ . Points $B$ and $C$ are chosen on these two lines such that $A B = A C$ . Determine the possible equations of the line $B C$ passing through the point $(1, 2)$ .

If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-

The two lines $x = a y + b, z = cy + d$ ; and $x = a^{'} y + b^{'}, z = c^{'} y + d^{'}$ are perpendicular to each other if

If the angle between the lines, $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{5 - x}{- 2} = \frac{7 y - 14}{p} = \frac{z - 3}{4}$ is $cos^{- 1} (\frac{2}{3})$ , then $p$ is equal to :

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2$ and $l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2}$ and $l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4}$ is:

The equation of one of the straight lines which passes through the point (1,3) and makes an angles $tan^{- 1} (2)$ with the straight line, $y + 1 = 32 x$ is