2d geometry system practice · Easy

Option C ($\frac{1}{a}+\frac{1}{b}=1$) is correct because it represents the condition derived from setting the determinant of the collinear points to zero.

Option A (-1) is correct because the product of the slopes of the two perpendicular lines is -1, which allows us to solve for h.

Option D (only one integral value of $k$) is correct because the area equation simplifies to a quadratic, yielding only one integer solution for k.

Option D ($x^2 + y^2 - a^2 = 0$) is correct because the equation represents a circle with radius 'a' centered at the origin, derived from the condition $PA = kPB$.

Option D (0) is correct because the points are collinear, resulting in a triangle with zero area, which can be shown by determinant properties.

Option B ($x^2+y^2=C-a^2$) is correct because it represents the locus of a point whose sum of squared distances from two fixed points is constant.

Option A $\left(2, \frac{{1}}{{2}}\right)$ is correct because after finding the value of k and the vertices, the orthocenter is calculated using properties of circumcenter and centroid.

Option A (independent of a) is correct because the area of the triangle simplifies to a constant value (1) regardless of the value of 'a'.

Option B (equilateral) is correct because a triangle with integral coordinates cannot be equilateral due to area considerations involving $\sqrt{3}$.

Option D ($\frac{8}{\sqrt{3}}$) is correct because it represents the area of the equilateral triangle calculated using the distance from point A to the origin and the properties of equilateral triangles.

Option B ($\frac{1}{\sqrt{2}}$) is correct because the given points form a right-angled triangle, and using the distance formula and trigonometric identities, we can find the value of $\left|\sin \left(\frac{\alpha-\beta}{2}\right)\right|$.

Option B ($x_1 x_2+y_1 y_2$) is correct because it represents the dot product of the vectors OA and OB, which is equal to |OA||OB|cos(∠AOB).

Option A ($a b^{\prime}=a^{\prime} b$) is correct because the condition for collinearity leads to the proportionality of the coordinates.

Option A (0) is correct because substituting x=3 and y=4 into the third equation and solving for 'a' yields a=0.

Option C is correct. If two vertices have integral coordinates, the third vertex of an equilateral triangle must have at least one irrational coordinate due to the area involving sqrt(3).

Option C (2) is correct because the value of p is found by setting the determinant of the matrix formed by the points to zero, indicating collinearity.

Option A ($(0,-1)$) is correct because the possible fourth vertices of the parallelogram are (-11, -8), (7, 10), and (-1, 0), making (0, -1) the only impossible vertex.

Option D (none of these) is correct because the calculated value of 'a' does not match any of the provided options.

Option C ($\left(a t_1^2, 2 a t_1\right)$) is correct because the orthocenter of $\triangle O'BC$ is simply point A, given that O' is the orthocenter of triangle ABC.

Option B (Equilateral triangle) is correct because the distances between all three points are equal, indicating an equilateral triangle.

Option B ($x^{2}-3 y^{2}=0$) is correct because the distance from the point to the x-axis ($|y|$) is half its distance from the origin ($\sqrt{{x^2+y^2}}$), leading to the equation $|y| = \frac{{1}}{{2}} \sqrt{{x^2+y^2}}$, which simplifies to $x^{2}-3 y^{2}=0$.

The correct answer is D (($x-4)^2+(y-5)^2=25$) because the incenter (1,1) satisfies this equation.

Option B ((7,2)) is correct because it satisfies both the conditions PA=PB and Area of triangle PAB=10.

Option C (7) is correct because substituting k=7 into the determinant equation for collinearity makes the determinant equal to zero.

Option C (Breadth $\geq 20$) is correct because the perimeter is $2(l+b)$ where $l=3b$, and the minimum perimeter is 160 cm, leading to $b \geq 20$.

Option A ($y^{2}=4 a x$) is correct because the distance from the point $(x,y)$ to $(a,0)$ being equal to its distance from the line $x+a=0$ leads to the equation $(x-a)^2 + y^2 = (x+a)^2$, which simplifies to $y^{2}=4 a x$.

Option D ($13 / 6$) is correct because the area of the triangle is calculated using the determinant formula and absolute value.

Option C ($a=1$ or $23 / 3$) is correct because the area of a triangle can be calculated using determinant formula and solving for 'a'.

Option A (-4, -3) is correct because using the centroid and circumcenter coordinates, we can find the orthocenter coordinates using the property that the centroid divides the line joining the orthocenter and circumcenter in the ratio 2:1.

Option C ($(a \Sigma \cos \theta_1, a \Sigma \sin \theta_1)$) is correct because the orthocenter of a triangle inscribed in a circle with center at the origin is given by the sum of the vertices.