An object was launched from a place $P$ at a constant speed to hit a target. At the $15$th second it was $1400$ m away from the target, and at the $18$th second $800$ m away. Find (i) the distance between the place and the target (ii) the distance covered by it in $15$ seconds (iii) the time taken to hit the target.

An object was launched from a place P at a constant speed to hit a target. At the 15th second it was 1400 m aw

Exercise 6.2 · Q6

P

(i) the distance between the place and the target

(ii) the distance covered by it in

15

seconds

(iii) the time taken to hit the target.

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Concept understanding — Straight Lines — Forms of the Equation

The general (linear) equation of a straight line is $ax+by+c=0$ , where $a,b$ are not both zero; the set of solutions of any such equation is a straight line in the plane. Because dividing through by $b$ (or $a$ ) removes one constant, every line's equation genuinely contains only two independent arbitrary constants — so exactly two independent pieces of information (two points, or a point and a slope, or two intercepts, etc.) are enough to pin a line down uniquely.

Slope. The angle of inclination $\theta$ of a line is the angle it makes with the $x$ -axis, measured counter-clockwise; the slope $m=\tan\theta$ (undefined when $\theta=\pi/2$ , i.e. for a vertical line). Equivalently, through two points $(x_1,y_1),(x_2,y_2)$ with $x_1\ne x_2$ , $m=\dfrac{y_2-y_1}{x_2-x_1}$ ; from the general form, $m=-a/b$ ( $b\ne0$ ). Three points are collinear exactly when the slope of any one pair equals the slope of another pair sharing a point.

Intercepts. The $x$ -intercept is where a line meets the $x$ -axis ( $y=0$ ); the $y$ -intercept is where it meets the $y$ -axis ( $x=0$ ). ( $x=0$ is itself the equation of the $y$ -axis; $y=0$ is the equation of the $x$ -axis.)

The six forms (two conditions each, all interconvertible by algebra):

Data given	Equation
Slope $m$ , $y$ -intercept $b$	$y=mx+b$
Slope $m$ , point $(x_1,y_1)$	$y-y_1=m(x-x_1)$
Two points $(x_1,y_1),(x_2,y_2)$	$\dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$ , equivalently $\begin{vmatrix}x-x_1 & y-y_1\\ x_2-x_1 & y_2-y_1\end{vmatrix}=0$
$x$ -intercept $a$ , $y$ -intercept $b$ (both $\ne0$ )	$\dfrac{x}{a}+\dfrac{y}{b}=1$
Normal length $p$ , angle $\alpha$ of the normal with the $x$ -axis	$x\cos\alpha+y\sin\alpha=p$
Parametric, through $(x_1,y_1)$ at inclination $\theta$ , parameter $r$ = signed distance from $(x_1,y_1)$	$\dfrac{x-x_1}{\cos\theta}=\dfrac{y-y_1}{\sin\theta}=r$

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Model the distance-from-target as a linear (constant-speed) function of time using the two given data points, then evaluate it at $t=0$ (launch) and solve for $d=0$ (impact).

Constant speed towards a fixed target means the distance remaining, $d$ , decreases linearly with time $t$ : $d=mt+c$ .

Step 1. Set up the two known points. At $t=15$ , $d=1400$ ; at $t=18$ , $d=800$ . So the two points on the line (with $t$ as $x$ , $d$ as $y$ ) are $(15,1400)$ and $(18,800)$ .

Step 2. Find the slope.

m=\frac{800-1400}{18-15}=\frac{-600}{3}=-200

(the distance shrinks by $200$ m every second — this is the object's speed.)

Step 3. Write the equation (point-slope through $(15,1400)$ ).

d-1400=-200(t-15) \;\Rightarrow\; d=1400-200t+3000=4400-200t

Step 4. Part (i) — distance between $P$ and the target. This is the distance at the moment of launch, $t=0$ :

$d(0)=4400-200(0)=4400\text{ m}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL1 markMCQ
$(2, 3)$
(a) $(-2, -3)$
(b) $(-3, -2)$
(c) $(3, 2)$
(d) $(-3, 2)$
›Reveal solutionSolution

Reflecting $(x,y)$ in the line $y=-x$ gives $(-y,-x)$ , so $(2,3)\to(-3,-2)$ .

The line $y=-x$ passes through the origin at $135^\circ$ to the x-axis. Reflection in this line swaps the coordinates and negates both: the general rule is $(x,y)\mapsto(-y,-x)$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$(2, 3)$
(a) $\alpha + 3\beta = 11$
(b) $\alpha + 2\beta = 7$
(c) $3\alpha + \beta = 11$
(d) $3\alpha + \beta = 9$
›Reveal solutionSolution

Find the equation of the line through the two given points and read off the condition satisfied by any point on it.

Slope $=\dfrac{4-3}{-1-2}=\dfrac{1}{-3}=-\dfrac13$ .

Equation: $y-3=-\dfrac13(x-2)\Rightarrow 3(y-3)=-(x-2)\Rightarrow 3y-9=-x+2\Rightarrow x+3y=11$ . …

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CBSE 2025Set ANNUAL1 markMCQ
$(1, -1)$
(a) $4x+3y+7=0$
(b) $4x-3y-7=0$
(c) $3x+4y+7=0$
(d) $3x+4y-7=0$
›Reveal solutionSolution

Perpendicular slopes are negative reciprocals; use the point-slope form with the given point.

Rewrite $3x+4y=6$ as $y=-\dfrac34x+\dfrac64$ , so its slope is $-\dfrac34$ .

A line perpendicular to it has slope $\dfrac43$ (negative reciprocal).

Using point-slope form through $(1,-1)$ : …

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CBSE 2020Set ANNUAL1 markMCQ
$x - y + 5 = 0$
(a) $x - y - 5 = 0$
(b) $x + y - 5 = 0$
(c) $x + y + 5 = 0$
(d) $x + y + 10 = 0$
›Reveal solutionSolution

Find where the given line meets the $y$ -axis, then draw the perpendicular there.

The given line $x-y+5=0$ , i.e. $y=x+5$ , has slope $1$ . It meets the $y$ -axis where $x=0$ : $y=0+5=5$ , so the point of intersection is $(0,5)$ .

…

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CBSE 2019Set ANNUAL1 markMCQ
$\dfrac{x}{a}-\dfrac{y}{b}=0$
(a) $a=b$
(b) only for $a=1, b=1$
(c) $a>b$
(d) $a<b$
›Reveal solutionSolution

Writing $\dfrac{x}{a}-\dfrac{y}{b}=0$ as $y=\dfrac{b}{a}x$ shows the slope is $b/a$ ; setting this equal to 1 gives $a=b$ .

$\dfrac{x}{a}-\dfrac{y}{b}=0 \Rightarrow \dfrac{y}{b}=\dfrac{x}{a} \Rightarrow y = \dfrac{b}{a}x$ .

…

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CBSE 2019Set ANNUAL1 markMCQ
$(2, 3)$
(a) $\alpha+3\beta=11$
(b) $3\alpha+\beta=11$
(c) $\alpha+2\beta=7$
(d) $3\alpha+\beta=9$
›Reveal solutionSolution

The slope of the line through $(2,3)$ and $(-1,4)$ is $-1/3$ ; writing its equation and substituting $(\alpha,\beta)$ gives $\alpha+3\beta=11$ .

Slope $m = \dfrac{4-3}{-1-2} = \dfrac{1}{-3} = -\dfrac{1}{3}$ .

Using point-slope form through $(2,3)$ : $y-3 = -\dfrac{1}{3}(x-2)$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Which of the following has the greatest y-intercept in magnitude?$
(a) $3x + 4y = 5$
(b) $2x + 3y = 4$
(c) $4x + 5y = 6$
(d) $x + 2y = 3$
›Reveal solutionSolution
Computing $y$ -intercept $=c/b$ for each line, $x+2y=3$ gives the largest magnitude, $1.5$ .

For $ax+by=c$ , setting $x=0$ gives the $y$ -intercept $y=c/b$ .

$3x+4y=5 \Rightarrow y\text{-intercept} = 5/4 = 1.25$

$2x+3y=4 \Rightarrow y\text{-intercept} = 4/3 \approx 1.33$

$4x+5y=6 \Rightarrow y\text{-intercept} = 6/5 = 1.2$ …
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Q.An object was launched from a place PPP at a constant speed to hit a target. At the 151515th second it was 140014001400 m away from the target, and at the 181818th second 800800800 m away. Find

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Q.An object was launched from a place $P$ at a constant speed to hit a target. At the $15$ th second it was $1400$ m away from the target, and at the $18$ th second $800$ m away. Find

Unlock everything free for 14 days