IPMAT Indore Straight Lines — PYPs

The Setup: We're dealing with homogenous equations of degree 2, which represent pairs of straight lines passing through the origin. The equations are $2x^2+axy+3y^2=0$ and $2x^2+bxy-3y^2=0$. One line coincides, and the other lines are perpendicular. This is pure coordinate geometry boss-fight level. Let the slope of the shared line be $m$, and the perpendicular ones be $m_1$ and $m_2$.
Step 1: Translate the given equations to quadratic form in terms of slope ($y=mx$).
For the first pair, dividing by $x^2$ gives:

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

The product of the roots (slopes) is $m \cdot m_1 = \frac{2}{3} \implies m_1 = \frac{2}{3m}$.
Step 2: Repeat for the second pair.

$$-3m^2 + bm + 2 = 0$$

The product of slopes here is $m \cdot m_2 = -\frac{2}{3} \implies m_2 = -\frac{2}{3m}$.
Step 3: Use the perpendicularity condition for the non-coinciding lines. Their slopes multiply to $-1$:

$$m_1 \cdot m_2 = -1$$

$$\left(\frac{2}{3m}\right) \left(-\frac{2}{3m}\right) = -1 \implies -\frac{4}{9m^2} = -1 \implies m^2 = \frac{4}{9}$$

Step 4: Set up the algebra to find $(a^2+b^2)$. We know $m$ is a root for both equations, so both evaluate to 0:
$3m^2 + am + 2 = 0$ (Eq. 1)
$-3m^2 + bm + 2 = 0$ (Eq. 2)
Add them together: $(a+b)m + 4 = 0 \implies a+b = -\frac{4}{m}$
Subtract Eq. 2 from Eq. 1: $6m^2 + (a-b)m = 0 \implies a-b = -6m$
Step 5: Calculate the final target, $a^2 + b^2$. Use the identity $2(a^2+b^2) = (a+b)^2 + (a-b)^2$.
Square the sums:
$(a+b)^2 = \frac{16}{m^2} = \frac{16}{4/9} = 36$
$(a-b)^2 = 36m^2 = 36(\frac{4}{9}) = 16$

$$2(a^2+b^2) = 36 + 16 = 52 \implies a^2+b^2 = 26$$

Final Answer: 26

The Setup: A coordinate geometry freebie. The farthest point of a circle from an external point lies on the line through the point and the centre, on the far side - so the maximum distance is the distance to the centre plus the radius.
Step 1: Deconstruct the circle. From $x^2+y^2=4$, the centre is the origin $C(0,0)$ and the radius is $r=\sqrt{4}=2$.
Step 2: Distance from the point to the centre. Both $P(-5,0)$ and $C(0,0)$ lie on the x-axis, so the distance is just the gap in x-coordinates:

$$d=|-5-0|=5$$

Step 3: Check the point is outside. Since $d=5>r=2$, the point $P$ lies outside the circle. (For a point *inside*, the maximum would still be $d+r$, but the minimum formula flips - worth knowing which case you are in.)
Step 4: Add the radius. Travel from $P$ through the centre and out to the opposite edge:

$$\text{Max distance}=d+r=5+2=7$$

Concretely, the farthest point is $(2,0)$, and the distance from $(-5,0)$ to $(2,0)$ is indeed 7 - while the nearest point $(-2,0)$ sits only 3 away.

Final Answer: 7

The Setup: This is a coordinate geometry speedrun using intercept form. The line crosses the x-axis at some point $A(a,0)$ and the y-axis at $B(0,b)$. Saying $M$ divides the portion between the axes into two equal halves is just a fancy way of saying $M$ is the midpoint of segment $AB$.
Step 1: Apply the midpoint formula. The midpoint of $A(a,0)$ and $B(0,b)$ is $\left(\frac{a}{2},\frac{b}{2}\right)$, and we are told that point is $M(-5,4)$:

$$\frac{a}{2}=-5\implies a=-10$$

$$\frac{b}{2}=4\implies b=8$$

Step 2: Construct the line equation. Use the two-intercept form $\frac{x}{a}+\frac{y}{b}=1$:

$$\frac{x}{-10}+\frac{y}{8}=1$$

Step 3: Clear the fractions in one shot. Multiply every term by $80$ (a common multiple of 10 and 8 chosen to land directly on the answer's scale):

$$80\cdot\frac{x}{-10}+80\cdot\frac{y}{8}=80 \implies -8x+10y=80$$

Reordering gives $10y-8x=80$.
Step 4: Verify against all three known points. A midpoint slip is easy to make, so test the line on the two intercepts and $M$ itself:
* $A(-10,0)$: $10(0)-8(-10)=80$ ✓
* $B(0,8)$: $10(8)-8(0)=80$ ✓
* $M(-5,4)$: $10(4)-8(-5)=40+40=80$ ✓

All three sit on the line, and $M$ is genuinely halfway between $A$ and $B$. Running the other three options through $M(-5,4)$ fails every time, so the match is unique.

Final Answer: $10y - 8x = 80$

The Setup: Three straight lines are concurrent (they intersect at a single unified point). We must find the intersection of the first two lines and use it to solve for the missing constant $k$ in the third line.

Step 1: Find the intersection point of the first two lines.
Line 1: $x - y - 1 = 0 \implies x = y + 1$
Line 2: $2x + 3y - 12 = 0$
Substitute $x$ into the second equation:

$$2(y + 1) + 3y - 12 = 0$$

$$2y + 2 + 3y - 12 = 0$$

$$5y - 10 = 0 \implies y = 2$$

Substitute $y$ back to find $x$:

$$x = 2 + 1 = 3$$

The shared intersection point is $(3, 2)$.

Step 2: Solve for $k$.
For the three lines to be perfectly concurrent, Line 3 ($2x - 3y + k = 0$) must also pass precisely through $(3, 2)$.
Substitute $x = 3$ and $y = 2$ into the third equation:

$$2(3) - 3(2) + k = 0$$

$$6 - 6 + k = 0 \implies k = 0$$

Final Answer: 0

The Setup: We need to trace the clues to build the quadratic equation $f(x)$ entirely in terms of the variable $a$. Once we have the pure function, we can calculate its vertex to find its lowest possible floor, and then plug that into the final target expression.
Step 1: Use the $g(x)$ coordinate to crack the value of $b$. The line $g(x)$ passes through $(a,b)$, so we literally just plug those coordinates in.

$$b=-2(a) \implies b=-2a$$

Step 2: Use the x-intercept to crack $c$. The function $f(x)$ hits the axis at $x=-2$, meaning $f(-2)=0$.

$$a(-2)^2+b(-2)+c=0$$

$$4a-2b+c=0$$

Substitute $b=-2a$ into this equation:

$$4a-2(-2a)+c=0 \implies 4a+4a+c=0 \implies 8a+c=0 \implies c=-8a$$

Step 3: Construct the final quadratic and find its minimum.

$$f(x)=ax^2-2ax-8a=a(x^2-2x-8)$$

For an upward-facing parabola ($a>0$), the minimum value occurs exactly at its vertex $x=-b_{new}/(2a_{new})$.

$$x=\frac{-(-2a)}{2(a)}=\frac{2a}{2a}=1$$

Step 4: Calculate the minimum y-value of $f(x)$ by plugging in $x=1$.

$$f(1)=a(1^2)-2a(1)-8a=a-2a-8a=-9a$$

Step 5: Evaluate the final target expression $f(x)+9a+1$ at its minimum point.

$$(-9a)+9a+1=1$$

Final Answer: 1

The Setup: We are dealing with a coordinate geometry vibe check. We have a defined line segment $AB$ and a mystery line pulling up at exactly a 45-degree angle. We need to find the slope $m$ of this mystery line. The ultimate weapon here is the angle between two lines formula, which relies on the tangent function to bridge the slopes.

Step 1: Find the slope of the base line $AB$.
The standard formula for slope is $m_1 = \frac{y_2 - y_1}{x_2 - x_1}$.
Plug in the coordinates for $A(1,3)$ and $B(5,1)$:
$$m_1 = \frac{1 - 3}{5 - 1}$$
$$m_1 = \frac{-2}{4} = -\frac{1}{2}$$

Step 2: Deploy the angle formula.
The formula for the angle $\theta$ between two intersecting lines with slopes $m$ and $m_1$ is:
$$\tan(\theta) = \left|\frac{m - m_1}{1 + m \cdot m_1}\right|$$
We know $\theta = 45^\circ$, and $\tan(45^\circ)$ is straight up $1$. Let's plug everything into the mainframe:
$$1 = \left|\frac{m - \left(-\frac{1}{2}\right)}{1 + m\left(-\frac{1}{2}\right)}\right|$$

Step 3: Clean up the UI and set up the absolute value split.
Simplify the fraction inside the absolute value by multiplying the numerator and the denominator by $2$ to clear out the decimals/fractions:
$$1 = \left|\frac{2m + 1}{2 - m}\right|$$
Because it is wrapped in an absolute value, this equation branches into two separate timelines—one positive and one negative. We have to solve both to get all possible values for $m$.

Step 4: Solve Timeline 1 (The Positive Case).
Set the inside of the absolute value equal to $1$:
$$\frac{2m + 1}{2 - m} = 1$$
Cross-multiply to flatten it out:
$$2m + 1 = 2 - m$$
Move the $m$'s to the left and the constants to the right:
$$3m = 1$$
$$m = \frac{1}{3}$$

Step 5: Solve Timeline 2 (The Negative Case).
Set the inside of the absolute value equal to $-1$:
$$\frac{2m + 1}{2 - m} = -1$$
Cross-multiply:
$$2m + 1 = -1(2 - m)$$
$$2m + 1 = -2 + m$$
Subtract $m$ from both sides, and subtract $1$ from both sides:
$$m = -3$$

Final Answer: $-3, \frac{1}{3}$

The Setup: Coordinate Geometry boss rush. We have to find an intersection point, calculate a perpendicular slope, build a brand new line equation, and finally snipe its x-intercept.
Step 1: Find the crossover point of $x+2y=4$ and $2x+3y=6$.
Multiply the first equation by $2$: $2x+4y=8$.
Subtract the second equation from it:

$$(2x+4y)-(2x+3y)=8-6 \implies y=2$$

Plug $y=2$ back into the first equation:

$$x+2(2)=4 \implies x+4=4 \implies x=0$$

The lines intersect at $(0,2)$. This is our anchor point.
Step 2: Steal the slope from the reference line. The reference is $3x-y=2$, which rearranges to $y=3x-2$. The slope ($m$) is $3$.
Step 3: Calculate the perpendicular slope. Perpendicular lines have negative reciprocal slopes.

$$m_{\perp}=-\frac{1}{3}$$

Step 4: Build the final line equation using the point-slope form with our anchor point $(0,2)$ and new slope $-1/3$.

$$y-2=-\frac{1}{3}(x-0)$$

$$3y-6=-x \implies x+3y=6$$

Step 5: Find the x-intercept. The x-intercept always occurs exactly when $y=0$.

$$x+3(0)=6 \implies x=6$$

Final Answer: 6

The Setup: We are dropping into a coordinate geometry battle royale. We have three straight lines acting as boundaries, and we need to find the exact area of the safe zone (the triangle) they form. To do this, we need to find the three Points of Interest (POIs) where these lines intersect, then run the standard base-times-height formula.

Step 1: Identify the boundary lines.
Here is our loadout:
Line 1: $y=0$ (This is literally just the x-axis, the ground floor).
Line 2: $12x-5y=0$
Line 3: $3x+4y=7$

Step 2: Locate the first two vertices (the ground floor POIs).
Since Line 1 is $y=0$, finding where it crosses the other two lines is a free loot drop. Just plug $0$ in for $y$ in their equations.
Intersection of Line 1 and Line 2:
$$12x - 5(0) = 0$$
$$12x = 0 \implies x = 0$$
First vertex locked in: $(0, 0)$.

Intersection of Line 1 and Line 3:
$$3x + 4(0) = 7$$
$$3x = 7 \implies x = \frac{7}{3}$$
Second vertex locked in: $(\frac{7}{3}, 0)$.

Step 3: Locate the final vertex (the apex).
Now we find where Line 2 and Line 3 cross each other (the third-party fight).
From Line 2, let's isolate $12x$:
$$12x = 5y$$
Now, take Line 3 ($3x+4y=7$) and multiply the entire equation by $4$ to match the $12x$ term:
$$12x + 16y = 28$$
Substitute $5y$ in for $12x$:
$$5y + 16y = 28$$
$$21y = 28 \implies y = \frac{28}{21} = \frac{4}{3}$$
So the y-coordinate (which is the exact vertical height of our triangle) is $\frac{4}{3}$.
*(We do not even need the x-coordinate because we already have the base secured flat on the x-axis, saving us precious time).*

Step 4: Calculate the final area.
Because our triangle sits perfectly flat on the x-axis, the math is super clean.
Base length ($b$) = Distance from $(0, 0)$ to $(\frac{7}{3}, 0)$ = $\frac{7}{3}$.
Height ($h$) = The y-coordinate of our apex vertex = $\frac{4}{3}$.
Plug these into the classic area formula:
$$\text{Area} = \frac{1}{2} \cdot b \cdot h$$
$$\text{Area} = \frac{1}{2} \cdot \frac{7}{3} \cdot \frac{4}{3}$$
$$\text{Area} = \frac{28}{18}$$
Simplify the fraction by dividing the top and bottom by $2$ for the flawless victory:
$$\text{Area} = \frac{14}{9}$$

Final Answer: $\frac{14}{9}$

The Setup: A locus question. When a point moves so that the ratio of its distances to two fixed points stays constant, the path is completely determined by that constant - this is the Circle of Apollonius.
Step 1: State the condition. Let the fixed points be $A$ and $B$ with $AB=1$ m, and let $P$ be the cow. The condition is:

$$\frac{PA}{PB}=\frac{3}{2}$$

Step 2: Apply the Apollonius result. For $\frac{PA}{PB}=k$ with $k>0$:
* If $k=1$ the locus is the perpendicular bisector of $AB$ - a straight line, since equal distances means equidistant.
* If $k\neq 1$ the locus is a circle.

Here $k=\frac{3}{2}\neq 1$, so the cow traces a circle.
Step 3: See it concretely rather than just quoting the theorem. Put $A$ at the origin and $B$ at $(1,0)$, and let $P=(x,y)$. Then $PA^2=\tfrac{9}{4}PB^2$ gives:

$$4\left(x^2+y^2\right)=9\left((x-1)^2+y^2\right) \implies 5x^2+5y^2-18x+9=0$$

$$x^2+y^2-\frac{18}{5}x+\frac{9}{5}=0 \implies \left(x-\frac{9}{5}\right)^2+y^2=\left(\frac{6}{5}\right)^2$$

That is unmistakably a circle - centre $\left(1.8,\,0\right)$, radius $1.2$ m. The two points where it crosses the line $AB$ are the internal and external division points of $AB$ in the ratio $3:2$, namely $x=0.6$ and $x=3$, which is exactly the diameter $[0.6,\,3]$ of width $2.4$.
Note the circle is not centred on $AB$'s midpoint and does not pass through $A$ or $B$ - a useful check that this really is Apollonius and not the perpendicular bisector.

Final Answer: the cow moves in a circle