Class 9 · Advanced MathematicsNCERT 2026 Revised Syllabus
Chapter 4: Coordinate Geometry
Class IX CBSE Advanced Mathematics
Uses the Cartesian plane to solve quadrant and reflection problems, slope-based questions on parallel and perpendicular lines, intercept form of a straight line, and the area of triangles cut off by lines with the axes.
Find complete NCERT solutions for Class 9 Advanced Mathematics, Class IX CBSE Advanced Mathematics Chapter 4. Every question has a clear, step-by-step answer.
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Exercise 4.1(2 questions)
1
HardPoint
Point $P$ lies in the III quadrant. Where does its reflection $P\text{\u2019}$ across the y-axis lie? Where does the reflection of $P\text{\u2019}$ across the x-axis lie?
School Level Solution
Sol.
Solution:
$P$ is in the III quadrant $(-, -)$.Reflecting across the y-axis flips the sign of the x-coordinate → $P\text{\u2019}$ is in $(+, -)$ = IV quadrant.Reflecting $P\text{\u2019}$ across the x-axis flips the sign of the y-coordinate → $P\text{\u2019\text{\u2019}}$ is in $(+, +)$ = I quadrant.Answer: $\boxed{P\text{\u2019}\text{ lies in the IV quadrant, }P\text{\u2019\text{\u2019}}\text{ lies in the I quadrant}}$.2
HardPoint
Point $A(a, b)$ is reflected across the x-axis to give $B(a, -b)$. If $B$ is 8 units below $A$, what is the y-coordinate of $A$?
School Level Solution
Sol.
Solution:Distance from
$A$ to $B$ is:
$$b - (-b) = 8 \Rightarrow 2b = 8 \Rightarrow b = 4$$Answer: $\boxed{\text{y-coordinate of }A = 4}$.Exercise 4.2(3 questions)
1
HardPoint
Point $P(a, b)$ lies in the IV quadrant. Its distance from the y-axis is 3 more than its distance from the x-axis, and $ab = -28$. Find $a^2 + b$.
School Level Solution
Sol.
Solution:In the IV quadrant,
$a > 0$, $b < 0$.Distance from y-axis $= a$; distance from x-axis $= -b$.Given: $a = (-b) + 3 \Rightarrow a + b = 3$. Also $ab = -28$.From $a = 3 - b$:
$$(3-b)b = -28 \Rightarrow b^2 - 3b - 28 = 0 \Rightarrow (b-7)(b+4) = 0$$Since $b < 0$, $b = -4$, so $a = 3 - (-4) = 7$. Check: $ab = 7(-4) = -28$ ✓.$$a^2 + b = 49 + (-4) = 45$$Answer: $\boxed{a^2 + b = 45}$.2
HardPoint
Point $Q(k-4, 2k+7)$ lies strictly in the II quadrant and its distance from the y-axis is twice its distance from the x-axis. Find $k$ and $k^3 + 10$.
School Level Solution
Sol.
Solution:For II quadrant:
$k - 4 < 0$ and $2k + 7 > 0$ ⇒ $-3.5 < k < 4$.Distance from y-axis $= 4 - k$; distance from x-axis $= 2k + 7$.Given: $4 - k = 2(2k + 7)$:
$$4 - k = 4k + 14 \Rightarrow -10 = 5k \Rightarrow k = -2$$$k = -2$ lies in the valid range.$$k^3 + 10 = (-8) + 10 = 2$$Answer: $\boxed{k = -2\text{, }k^3 + 10 = 2}$.3
HardPoint
Point $P(3, -4)$ is reflected through the y-axis to $Q(a, b)$ and through the x-axis to $R(c, d)$. Find $ac - bd$.
School Level Solution
Sol.
Solution:Reflection through y-axis:
$Q(a, b) = (-3, -4)$.Reflection through x-axis: $R(c, d) = (-3, 4)$.$$ac - bd = (-3)(-3) - (-4)(4) = 9 + 16 = 25$$Answer: $\boxed{ac - bd = 25}$.Exercise 4.3(5 questions)
1
HardA line of slope
A line of slope $\frac{1}{2}$ passes through $A(x, 2)$, $B(3, 4)$ and $C(7, y)$. Find $x$ and $y$, and hence $x + y$.
School Level Solution
Sol.
Solution:Slope of
$AB$:
$$\frac{4-2}{3-x} = \frac{1}{2} \Rightarrow \frac{2}{3-x} = \frac{1}{2} \Rightarrow 4 = 3 - x \Rightarrow x = -1$$Slope of $BC$:
$$\frac{y-4}{7-3} = \frac{1}{2} \Rightarrow \frac{y-4}{4} = \frac{1}{2} \Rightarrow y - 4 = 2 \Rightarrow y = 6$$$$x + y = -1 + 6 = 5$$Answer: $\boxed{A(-1, 2)\text{, }C(7, 6)\text{, }x + y = 5}$.2
HardThe points
The points $P(2, 3)$, $Q(5, 7)$ and $R(13, k)$ are consecutive vertices of a rectangle. Find $k$.
School Level Solution
Sol.
Solution:In a rectangle, adjacent sides are perpendicular, so
$PQ \perp QR$.Slope of $PQ$:
$$\frac{7-3}{5-2} = \frac{4}{3}$$Slope of $QR$:
$$\frac{k-7}{13-5} = \frac{k-7}{8}$$Perpendicular condition:
$$\frac{4}{3} \times \frac{k-7}{8} = -1 \Rightarrow \frac{4(k-7)}{24} = -1 \Rightarrow k - 7 = -6 \Rightarrow k = 1$$Answer: $\boxed{k = 1}$.3
HardA line
A line $l$ through the origin intersects segment $AB$ where $A(2, 8)$ and $B(6, 2)$. Between which limits does the slope $m$ of $l$ lie?
School Level Solution
Sol.
Solution:Slope of
$OA$:
$$\frac{8}{2} = 4$$Slope of $OB$:
$$\frac{2}{6} = \frac{1}{3}$$As the line rotates from $OA$ to $OB$, its slope $m$ ranges between these two values:
$$\frac{1}{3} \leq m \leq 4$$Answer: $\boxed{\frac{1}{3} \leq m \leq 4}$.4
HardLine
Line $l_1$ passes through $A(1, k)$ and $B(k, 7)$; line $l_2$ passes through $C(-2, 4)$ and $D(1, 10)$. If $l_1 \parallel l_2$, find $k$ and $k^2 + 5$.
School Level Solution
Sol.
Solution:Slope of
$l_1$:
$$\frac{7-k}{k-1}$$Slope of $l_2$:
$$\frac{10-4}{1-(-2)} = \frac{6}{3} = 2$$Parallel lines have equal slopes:
$$\frac{7-k}{k-1} = 2 \Rightarrow 7 - k = 2k - 2 \Rightarrow 9 = 3k \Rightarrow k = 3$$$$k^2 + 5 = 9 + 5 = 14$$Answer: $\boxed{k = 3\text{, }k^2 + 5 = 14}$.5
HardLine
Line $p$ has slope 2. Line $q$ passes through $M(3a, 4)$ and $N(a, 5)$. If $p \perp q$, find $a$ and the coordinates of $M$ and $N$.
School Level Solution
Sol.
Solution:Slope of
$q$:
$$\frac{5-4}{a-3a} = \frac{1}{-2a}$$Perpendicular condition: $m_p \cdot m_q = -1$:
$$2 \times \left(-\frac{1}{2a}\right) = -1 \Rightarrow -\frac{1}{a} = -1 \Rightarrow a = 1$$$$M(3a, 4) = (3, 4), \quad N(a, 5) = (1, 5)$$Answer: $\boxed{a = 1\text{, }M(3, 4)\text{, }N(1, 5)}$.Exercise 4.4(4 questions)
1
HardThe line
The line $kx + 3y - 12 = 0$ forms a right-angled triangle with the x and y coordinate axes. If the total area of this triangle is 12 square units, find all possible values for the slope of this line.
School Level Solution
Sol.
Solution:x-intercept
$= \frac{12}{k}$; y-intercept $= 4$.Area $= \frac{1}{2} \times \frac{12}{k} \times 4 = \frac{24}{k} = 6 \Rightarrow |k| = 2 \Rightarrow k = \pm 2$.Slope of $kx + 3y - 12 = 0$ is $-\frac{k}{3}$:
- for $k = -2$: slope $= \frac{2}{3}$
- for $k = 2$: slope $= -\frac{2}{3}$Answer: $\boxed{\text{slope }= \pm \frac{2}{3}}$.2
HardThe line
The line $px + qy + r = 0$ forms an isosceles right triangle with the axes. What must be true about $p$ and $q$?
School Level Solution
Sol.
Solution:x-intercept
$= -\frac{r}{p}$; y-intercept $= -\frac{r}{q}$.For an isosceles right triangle with the axes, the intercepts must be equal in magnitude:
$$\left|-\frac{r}{p}\right| = \left|-\frac{r}{q}\right| \Rightarrow |p| = |q| \Rightarrow p^2 = q^2$$Answer: $\boxed{|p| = |q|\text{, i.e., }p^2 = q^2}$.3
HardGiven
Given $l_1: 3x - 5y + 10 = 0$ and $l_2: 5x + 3y + K = 0$. (a) Show $l_1 \perp l_2$. (b) If the x-intercept of $l_1$ equals the y-intercept of $l_2$, find $K$.
School Level Solution
Sol.
Solution:Slope of
$l_1$: $3x - 5y + 10 = 0 \Rightarrow y = \frac{3}{5}x + 2$, slope $= \frac{3}{5}$.Slope of $l_2$: $5x + 3y + K = 0 \Rightarrow y = -\frac{5}{3}x - \frac{K}{3}$, slope $= -\frac{5}{3}$.(a) Product of slopes:
$$\frac{3}{5} \times \left(-\frac{5}{3}\right) = -1 \Rightarrow l_1 \perp l_2$$(b) x-intercept of $l_1$: put $y = 0$: $3x + 10 = 0 \Rightarrow x = -\frac{10}{3}$.y-intercept of $l_2$: put $x = 0$: $3y + K = 0 \Rightarrow y = -\frac{K}{3}$.Given equal:
$$-\frac{10}{3} = -\frac{K}{3} \Rightarrow K = 10$$Answer: $\boxed{l_1 \perp l_2\text{ (proved); }K = 10}$.4
HardThe line
The line $kx - y + C = 0$ passes through $(3, 10)$, and the sum of its x-intercept and y-intercept equals the slope. Find the two possible lines.
School Level Solution
Sol.
Solution:
$kx - y + C = 0 \Rightarrow y = kx + C$, slope $= k$.Passes through $(3, 10)$: $10 = 3k + C \Rightarrow C = 10 - 3k$.x-intercept $= -\frac{C}{k}$; y-intercept $= C$.Given: $-\frac{C}{k} + C = k$. Substituting $C = 10 - 3k$ and simplifying:
$$4k^2 - 13k + 10 = 0 \Rightarrow k = \frac{13 \pm 3}{8} = 2 \text{ or } \frac{5}{4}$$For $k = 2$: $C = 4$; for $k = \frac{5}{4}$: $C = \frac{25}{4}$.Answer: $\boxed{2x - y + 4 = 0\text{ or }5x - 4y + 25 = 0}$.Exercise 4.5(5 questions)
1Write the equation of the line that crosses the y-axis at
Write the equation of the line that crosses the y-axis at $(0, -5)$ and rises 3 units for every 2 units to the right.
School Level Solution
Sol.
Solution:Rises 3 for every 2 right → slope
$= \frac{3}{2}$.y-intercept $= -5$.$$y = \frac{3}{2}x - 5$$Answer: $\boxed{y = \frac{3}{2}x - 5}$.2Write the equation of the line parallel to
Write the equation of the line parallel to $y = 5x - 12$ passing through $(0, 9)$.
School Level Solution
Sol.
Solution:Parallel lines have equal slope: slope
$= 5$.Through $(0, 9)$ → y-intercept $= 9$.$$y = 5x + 9$$Answer: $\boxed{y = 5x + 9}$.3
HardWrite the equation of the line perpendicular to
Write the equation of the line perpendicular to $y = \frac{1}{3}x + 4$ passing through the origin.
School Level Solution
Sol.
Solution:Slope of given line
$= \frac{1}{3}$, so the required perpendicular slope $= -3$.Through the origin → y-intercept $= 0$.$$y = -3x$$Answer: $\boxed{y = -3x}$.4Write the equation of the line through
Write the equation of the line through $(0, -6)$ and $(4, 10)$.
School Level Solution
Sol.
Solution:Slope:
$$\frac{10 - (-6)}{4 - 0} = \frac{16}{4} = 4$$y-intercept $= -6$.$$y = 4x - 6$$Answer: $\boxed{y = 4x - 6}$.5Express
Express $8x - y + 7 = 0$ in the form $y = mx + c$.
School Level Solution
Sol.
Solution:
$$8x - y + 7 = 0 \Rightarrow y = 8x + 7$$Answer: $\boxed{y = 8x + 7}$.Exercise 4.6(5 questions)
1
HardA line passes through
A line passes through $(2, 3)$ and cuts the axes such that the triangle area is 12. Find the x-intercept and y-intercept.
School Level Solution
Sol.
Solution:Using intercept form
$\frac{x}{a} + \frac{y}{b} = 1$:
$$\frac{2}{a} + \frac{3}{b} = 1, \quad \text{and} \quad \frac{1}{2}ab = 12 \Rightarrow ab = 24$$$$\frac{2}{a} + \frac{3}{b} = 1 \Rightarrow 2b + 3a = ab = 24 \Rightarrow 3a + 2b = 24$$From $ab = 24$: $b = \frac{24}{a}$. Substituting:
$$3a + \frac{48}{a} = 24 \Rightarrow 3a^2 - 24a + 48 = 0 \Rightarrow a^2 - 8a + 16 = 0 \Rightarrow (a-4)^2 = 0$$
$$a = 4, \quad b = 6$$Answer: $\boxed{\text{x-intercept }= 4\text{, y-intercept }= 6}$.2
HardA line passes through
A line passes through $(2, 2)$ and $a + b = 9$. Find $ab$.
School Level Solution
Sol.
Solution:Using intercept form:
$$\frac{2}{a} + \frac{2}{b} = 1 \Rightarrow \frac{2(a+b)}{ab} = 1 \Rightarrow \frac{2(9)}{ab} = 1 \Rightarrow ab = 18$$Answer: $\boxed{ab = 18}$.3
HardA line passes through
A line passes through $(3, 5)$ and $a + b = 0$. Find its equation.
School Level Solution
Sol.
Solution:
$a + b = 0 \Rightarrow b = -a$.Intercept form:
$$\frac{x}{a} - \frac{y}{a} = 1 \Rightarrow x - y = a$$Passes through $(3, 5)$: $3 - 5 = a \Rightarrow a = -2$.$$x - y = -2 \Rightarrow x - y + 2 = 0$$Answer: $\boxed{x - y + 2 = 0}$.4
Hard
A line forms a right triangle with the axes having area 24 and hypotenuse 10. Find its equation(s).
School Level Solution
Sol.
Solution:Let intercepts be
$a$ and $b$ (legs of the right triangle).$$\frac{1}{2}ab = 24 \Rightarrow ab = 48; \quad a^2 + b^2 = 100$$$$(a+b)^2 = a^2 + b^2 + 2ab = 100 + 96 = 196 \Rightarrow a + b = 14$$$$(a-b)^2 = a^2 + b^2 - 2ab = 100 - 96 = 4 \Rightarrow a - b = \pm 2$$Solving: $a = 8, b = 6$ or $a = 6, b = 8$.Answer: $\boxed{\frac{x}{8} + \frac{y}{6} = 1\text{ or }\frac{x}{6} + \frac{y}{8} = 1}$.5
HardA line passes through
A line passes through $(3, 2)$, $a + b = 12$ with both intercepts positive. Find its equation(s).
School Level Solution
Sol.
Solution:
$$\frac{3}{a} + \frac{2}{b} = 1, \quad \text{and} \quad b = 12 - a$$$$\frac{3}{a} + \frac{2}{12-a} = 1 \Rightarrow 3(12-a) + 2a = a(12-a)$$
$$36 - a = 12a - a^2 \Rightarrow a^2 - 13a + 36 = 0 \Rightarrow (a-9)(a-4) = 0$$$a = 9, b = 3$ or $a = 4, b = 8$.Answer: $\boxed{\frac{x}{9} + \frac{y}{3} = 1\text{ or }\frac{x}{4} + \frac{y}{8} = 1}$.