2d geometry system practice · Mixed difficulty

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.

Option B ((-1/10), (37/10)) is correct because it represents the coordinates of the foot of the perpendicular from the given point to the line.

Option D ($(a+1) x-(a-1) y=2 a$) is correct because the orthocenter lies on the line that satisfies the relation between circumcenter, centroid, and orthocenter.

Option B (A circle) is correct because substituting $h = \frac{{x_1^2 - 2}}{{3}}$ and $k = \frac{{y_1^2}}{{3}}$ into $h + k = 1$ results in the equation of a circle: $x^2 + y^2 = 5$.

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

Option B ($(x+3 y+1)(x+3 y-23)=0$) is correct because it represents the locus of point P such that the area of triangle PAB is 12 sq. units.

Option A (-1, 89) is correct because the area of the pentagon is calculated using the determinant formula and solving for t.

Option C (a=2, b=3) is correct because in a parallelogram, the midpoints of the diagonals coincide, allowing us to solve for a and b.

Option C ($a^{2}\left(m_{2}-m_{3}\right)\left(m_{3}-m_{1}\right)\left(m_{1}-m_{2}\right)$) is correct because it is the result of calculating the area of the triangle using the determinant formula.

Option A (5) is correct because substituting p=5 into the determinant equation for collinearity makes the determinant equal to zero.

Option A $(a, b)$ is correct because the centroid's coordinates are derived directly from Vieta's formulas and the given equation's coefficients.

Option A ($y=-1$) is correct because the product of the slopes of $AB$ and $AC$ must be -1 for the triangle to be right-angled at $A$.

Option B (a right angled triangle) is correct because the square of one side equals the sum of the squares of the other two sides, satisfying the Pythagorean theorem.

Option A (First) is correct because the orthocentre, calculated as the intersection of altitudes, lies in the first quadrant with coordinates $\left(\frac{{3}}{{7}}, \frac{{22}}{{7}}\right)$.

Option D (-1, -5) is correct because by equating the area of the triangle to the given value and solving the resulting exponential equation, we find x = -1 or x = -5.

Option A (3,3) is correct because the centroid is calculated by averaging the x and y coordinates of the vertices.

Option A ($b x \pm a y=0$) is correct because it represents the locus of point C as A and B slide along the axes, maintaining the right angle at C.

Option D ($x^{2}+y^{2}-4=0$) is correct because the sum of the squares of the distances from $(x,y)$ to $(2,0)$ and $(-2,0)$ being equal to the square of the distance between $(2,0)$ and $(-2,0)$ leads to the equation $(x-2)^2 + y^2 + (x+2)^2 + y^2 = 16$, which simplifies to $x^{2}+y^{2}=4$ or $x^{2}+y^{2}-4=0$.

Option C ((5,420°)) is correct because adding multiples of 360° to the angle in polar coordinates results in an equivalent point.

Option A ((-4/3, -4/3)) is correct because it represents the intersection point of the altitudes of the triangle formed by the given lines.

Option A ($x-y=0$) is correct because it represents the locus of point P equidistant from points A and B.

Option C ($c=-4$) is correct because the centroid of the triangle lies on the x-axis, implying the y-coordinate of the centroid is zero, leading to $c=-4$.

Option A (2) is correct because it is the area of the triangle calculated using the determinant formula with the given vertices.

Option B (Parallelogram) is correct. The quadrilateral is a parallelogram because opposite sides are equal in length, but the diagonals are not equal, indicating it is not a rectangle or square.

Option C (3:4) is correct because the line divides the segment in the ratio 3:4, which is found by substituting the section formula into the line's equation.

Option B ($(1, \sqrt{{3}})$) is correct because it represents the Cartesian coordinates corresponding to the given polar coordinates.

Option A is correct because the centroid divides the line joining the circumcentre and orthocentre in the ratio 2:1.

Option D (parallelogram) is correct because opposite sides are parallel, but not all sides are equal, and the angles are not necessarily right angles.