If one of the lines given by the equation 2x2+axy+3y2=0 coincides with one of those given by 2x2+bxy−3y2=0 and the other lines represented by them are perpendicular then a2+b2=
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The Setup: We're dealing with homogenous equations of degree 2, which represent pairs of straight lines passing through the origin. The equations are 2x2+axy+3y2=0 and 2x2+bxy−3y2=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 m1 and m2.
Step 1: Translate the given equations to quadratic form in terms of slope (y=mx).
For the first pair, dividing by x2 gives:
3m2+am+2=0
The product of the roots (slopes) is m⋅m1=32⟹m1=3m2.
Step 2: Repeat for the second pair.
−3m2+bm+2=0
The product of slopes here is m⋅m2=−32⟹m2=−3m2.
Step 3: Use the perpendicularity condition for the non-coinciding lines. Their slopes multiply to −1:
m1⋅m2=−1(3m2)(−3m2)=−1⟹−9m24=−1⟹m2=94Step 4: Set up the algebra to find (a2+b2). We know m is a root for both equations, so both evaluate to 0:
3m2+am+2=0 (Eq. 1)
−3m2+bm+2=0 (Eq. 2)
Add them together: (a+b)m+4=0⟹a+b=−m4
Subtract Eq. 2 from Eq. 1: 6m2+(a−b)m=0⟹a−b=−6mStep 5: Calculate the final target, a2+b2. Use the identity 2(a2+b2)=(a+b)2+(a−b)2.
Square the sums:
(a+b)2=m216=4/916=36(a−b)2=36m2=36(94)=162(a2+b2)=36+16=52⟹a2+b2=26Final Answer: 26
The maximum distance between the point (−5,0) and a point on the circle x2+y2=4 is
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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 x2+y2=4, the centre is the origin C(0,0) and the radius is r=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∣=5Step 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:
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 equation of the straight line passing through the point M(−5,4), such that the portion of it between the axes is divided by the point M into two equal halves, is
A10y−8x=80
B8y+10x=80
C10y+8x=80
D8y+10x+80=0
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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 (2a,2b), and we are told that point is M(−5,4):
2a=−5⟹a=−102b=4⟹b=8Step 2: Construct the line equation. Use the two-intercept form ax+by=1:
−10x+8y=1Step 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⋅−10x+80⋅8y=80⟹−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 value of k for which the following lines are concurrent is
x−y−1=02x+3y−12=02x−3y+k=0
A1
B-1
C0
D12
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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⟹x=y+1
Line 2: 2x+3y−12=0
Substitute x into the second equation:
2(y+1)+3y−12=02y+2+3y−12=05y−10=0⟹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=06−6+k=0⟹k=0Final Answer: 0
Consider the polynomials f(x)=ax2+bx+c, where a>0,b,c are real, g(x)=−2x. If f(x) cuts the x-axis at (−2,0) and g(x) passes through (a,b), then the minimum value of f(x)+9a+1 is
A0
B1
C2
D3
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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)⟹b=−2aStep 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=04a−2b+c=0
Substitute b=−2a into this equation:
4a−2(−2a)+c=0⟹4a+4a+c=0⟹8a+c=0⟹c=−8aStep 3: Construct the final quadratic and find its minimum.
f(x)=ax2−2ax−8a=a(x2−2x−8)
For an upward-facing parabola (a>0), the minimum value occurs exactly at its vertex x=−bnew/(2anew).
x=2(a)−(−2a)=2a2a=1Step 4: Calculate the minimum y-value of f(x) by plugging in x=1.
f(1)=a(12)−2a(1)−8a=a−2a−8a=−9aStep 5: Evaluate the final target expression f(x)+9a+1 at its minimum point.
(−9a)+9a+1=1Final Answer: 1
Let A(1,3) and B(5,1) be two points. If a line with slope m intersects AB at an angle of 45∘, then the possible values of m are
A7,71
B3,31
C−3,31
D5,−51
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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 m1=x2−x1y2−y1.
Plug in the coordinates for A(1,3) and B(5,1):
m1=5−11−3m1=4−2=−21Step 2: Deploy the angle formula.
The formula for the angle θ between two intersecting lines with slopes m and m1 is:
tan(θ)=1+m⋅m1m−m1
We know θ=45∘, and tan(45∘) is straight up 1. Let's plug everything into the mainframe:
1=1+m(−21)m−(−21)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=2−m2m+1
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:
2−m2m+1=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=1m=31Step 5: Solve Timeline 2 (The Negative Case).
Set the inside of the absolute value equal to −1:
2−m2m+1=−1
Cross-multiply:
2m+1=−1(2−m)2m+1=−2+m
Subtract m from both sides, and subtract 1 from both sides:
m=−3Final Answer:−3,31
The x-intercept of the line that passes through the intersection of the lines x+2y=4 and 2x+3y=6, and is perpendicular to the line 3x−y=2 is
A2
B0.5
C4
D6
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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⟹y=2
Plug y=2 back into the first equation:
x+2(2)=4⟹x+4=4⟹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⊥=−31Step 4: Build the final line equation using the point-slope form with our anchor point (0,2) and new slope −1/3.
y−2=−31(x−0)3y−6=−x⟹x+3y=6Step 5: Find the x-intercept. The x-intercept always occurs exactly when y=0.
x+3(0)=6⟹x=6Final Answer: 6
The area of the triangle, formed by the straight lines y=0, 12x−5y=0, and 3x+4y=7 is
A2735
B914
C928
D5435
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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=7Step 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)=012x=0⟹x=0
First vertex locked in: (0,0).
Intersection of Line 1 and Line 3:
3x+4(0)=73x=7⟹x=37
Second vertex locked in: (37,0).
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=2821y=28⟹y=2128=34
So the y-coordinate (which is the exact vertical height of our triangle) is 34.
*(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 (37,0) = 37.
Height (h) = The y-coordinate of our apex vertex = 34.
Plug these into the classic area formula:
Area=21⋅b⋅hArea=21⋅37⋅34Area=1828
Simplify the fraction by dividing the top and bottom by 2 for the flawless victory:
Area=914Final Answer:914
Two points on a ground are 1 m apart. If a cow moves in the field in such a way that its distance from the two points is always in ratio 3:2 then
Athe cow moves in a straight line
Bthe cow moves in a circle
Cthe cow moves in a parabola
Dthe cow moves in a hyperbola
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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:
PBPA=23Step 2: Apply the Apollonius result. For PBPA=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=1 the locus is a circle.
Here k=23=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 PA2=49PB2 gives:
4(x2+y2)=9((x−1)2+y2)⟹5x2+5y2−18x+9=0x2+y2−518x+59=0⟹(x−59)2+y2=(56)2
That is unmistakably a circle - centre (1.8,0), 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